Sequences and Series

Given

$ \sum\limits_{r=1}^{25}\left(\frac{r}{r^4+r^2+1}\right)=\frac{p}{q} $

The expression $r^4+r^2+1$ can be factored using a completion of squares technique:

$ r^4+r^2+1=\left(r^4+2 r^2+1\right)-r^2=\left(r^2+1\right)^2-r^2 $

Using the difference of squares formula, $a^2-b^2=(a-b)(a+b)$ :

$ r^4+r^2+1=\left(r^2-r+1\right)\left(r^2+r+1\right) $

Split into Partial Fractions

Now, let's rewrite the general term $T_r$ :

$ T_r=\frac{r}{\left(r^2-r+1\right)\left(r^2+r+1\right)} $

We can express the numerator $r$ in terms of the factors in the denominator:

$ \left(r^2+r+1\right)-\left(r^2-r+1\right)=2 r $

So, $r=\frac{1}{2}\left[\left(r^2+r+1\right)-\left(r^2-r+1\right)\right]$. Substituting this back:

$ T_r=\frac{1}{2}\left[\frac{\left(r^2+r+1\right)-\left(r^2-r+1\right)}{\left(r^2-r+1\right)\left(r^2+r+1\right)}\right] $

$ T_r=\frac{1}{2}\left[\frac{1}{r^2-r+1}-\frac{1}{r^2+r+1}\right] $

Evaluate the Telescoping Sum

Let $f(r)=\frac{1}{r^2-r+1}$.

Note that $f(r+1)=\frac{1}{(r+1)^2-(r+1)+1}=\frac{1}{r^2+2 r+1-r-1+1}=\frac{1}{r^2+r+1}$.

Our sum becomes : $ S=\frac{1}{2} \sum\limits_{r=1}^{25}[f(r)-f(r+1)] $

Expanding this:

• For $r=1$ : $f(1)-f(2)$

• For $r=2: f(2)-f(3)$

• For $r=3: f(3)-f(4)$

• …

• For $r=24: f(24)-f(25)$

• For $r=25: f(25)-f(26)$

Result of summation: $\sum\limits_{r=1}^{25}[f(r)-f(r+1)]=f(1)-f(26)$

So,

$ S=\frac{1}{2}[f(1)-f(26)] $

Calculate the Final Value

• $f(1)=\frac{1}{1^2-1+1}=1$

$\begin{aligned} & f(26)=\frac{1}{25^2+25+1}=\frac{1}{625+25+1}=\frac{1}{651} \\ & S=\frac{1}{2}\left[1-\frac{1}{651}\right]=\frac{1}{2}\left(\frac{651-1}{651}\right)=\frac{650}{2 \times 651} \\ & S=\frac{325}{651}\end{aligned}$

$325=(5)^2 \times 13$ and $651=3 \times 7 \times 31$ they share no common factor $\operatorname{gcd}(325,651)=1$

So $p=325$ and $q=651

$\therefore $ p+q=325+651=976$

Product of first three terms is 27.

Let the terms be $a / r, a, a r$.

$(a / r)(a)(a r)=27 \Rightarrow a^3=27 \Rightarrow a=3$

Sum $S=\frac{3}{r}+3+3 r=3\left(r+\frac{1}{r}+1\right)$.

Find the range of the sum

The sum $S=3\left(\frac{1}{r}+1+r\right)$.

Using the Arithmetic Mean-Geometric Mean (AM-GM) inequality for $r+\frac{1}{r}$ :

If $r>0$, then $r+\frac{1}{r} \geq 2$, so $S \geq 3(2+1)=9$.

If $r<0$, then $r+\frac{1}{r} \leq-2$, so $S \leq 3(-2+1)=-3$.

The possible values for the sum are $(-\infty,-3] \cup[9, \infty)$.

This is expressed as $\mathbb{R}-(-3,9)$.

Thus, $a=-3$ and $b=9$.

The final value is $a^2+b^2=(-3)^2+9^2=9+81=\mathbf{9 0}$.

$ \begin{aligned} & a=b-d, c=b+d, \Rightarrow b=\frac{1}{3} \Rightarrow 4 b^4=a^2 c^2 \\ & 4 b^4=[(b-d)(b+d)]^2 \\ & \frac{4}{81}=\left(\frac{1}{9}-d^2\right)^2 \Rightarrow \frac{4}{81}=\frac{1}{81}-\frac{2 d^2}{9}+d^4 \Rightarrow d^4-\frac{2 d^2}{9}-\frac{1}{27}=0 \Rightarrow 27 d^4-6 d^2-1=0 \\ & d^2=1 / 3 \Rightarrow d=+\frac{1}{\sqrt{3}}(\text { as } \mathrm{a}>\mathrm{b}>\mathrm{c}) \\ & 9\left(a^2+b^2+c^2\right)=9\left[\left(\frac{1}{3}-\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}+\frac{1}{\sqrt{3}}\right)^2\right]=9\left[\frac{1}{3}+\frac{2}{3}\right]=3+6=9 \end{aligned} $

$ \begin{aligned} & a_{n+1}=\frac{1}{2} a_n+\frac{1}{(n+1)^2}-\frac{\left((n+1)^2-n^2\right)}{n^2(n+1)^2} \\ & a_{n+1}=\frac{a_n}{2}+\frac{2}{(n+1)^2}-\frac{1}{n^2} \\ & a_{n+1}-\frac{2}{(n+1)^2}=\frac{1}{2}\left(a_n-\frac{2}{n^2}\right) \\ & \text { Let } b_n=a_n \frac{-2}{n^2} \end{aligned} $

$ \begin{aligned} &\text { then }\langle b\rangle \text { is geometric progression with ratio }=\frac{1}{2}\\ &\begin{aligned} & b_1=a_1-\frac{2}{1}=1-2=(-1) \\ & \sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)=\sum_{n=1}^{\infty} b_n=\frac{(-1)}{1-\left(\frac{1}{2}\right)}=\frac{-1}{1 / 2}=-2 \\ & \Rightarrow\left|\sum_{n=1}^{\infty}\left(a_n-\frac{2}{n^2}\right)\right|=2 \end{aligned} \end{aligned} $

$\begin{aligned} & \frac{4.1}{1+4.1^4}+\frac{4.2}{1+4.2^4}+\frac{4.3}{1+4.3^4}+\ldots . \\ & T_r=\frac{4 r}{1+4 r^4}=\frac{4 r}{4 r^4+4 r^2+1-4 r^2} \\ & =\frac{4 r}{\left(2 r^2+1\right)^2-(2 r)^2} \\ & T_r=\frac{4 r}{\left(2 r^2-2 r+1\right)\left(2 r^2+2 r+1\right)} \end{aligned}$

$\begin{aligned} & T_r=\frac{\left(2 r^2+2 r+1\right)-\left(2 r^2-2 r+1\right)}{\left(2 r^2-2 r+1\right)\left(2 r^2+2 r+1\right)} \\ & T_r=\left(\frac{1}{r^2+(r-1)^2}-\frac{1}{r^2+(r+1)^2}\right) \\ & \sum_{r=1}^{10} T_r=\left(\frac{1}{0^2+1^2}-\frac{1}{1^2+2^2}+\frac{1}{1^2+2^2}-\frac{1}{2^2+3^2}+\ldots .\right. \\ & \frac{1}{9^2+10^2}-\frac{1}{10^2+11^2} \\ & =1-\frac{1}{221} \\ & =\frac{220}{221} \\ & \therefore \quad m+n=220+221 \\ & \quad=441 \end{aligned}$

$\mathrm{a}_1+\mathrm{a}_5+\mathrm{a}_{10}+\ldots \ldots+\mathrm{a}_{2020}+\mathrm{a}_{2024}=2233$

In an A.P. the sum of terms equidistant from ends is equal.

$\begin{aligned} & a_1+a_{2024}=a_5+a_{2020}=a_{10}+a_{2015} \ldots \ldots \\ & \Rightarrow 203 \text { pairs } \\ & \Rightarrow 203\left(a_1+a_{2024}\right)=2233 \end{aligned}$

Hence,

$\begin{aligned} & \mathrm{S}_{2024}=\frac{2024}{2}\left(\mathrm{a}_1+\mathrm{a}_{2024}\right) \\ = & 1012 \times 11 \\ = & 11132 \end{aligned}$

$\begin{aligned} & \frac{\mathrm{n}}{2}(2 \mathrm{a}+(\mathrm{n}-1) 6)=(\mathrm{n}-2) \cdot 180^{\circ} \\ & \mathrm{an}+3 \mathrm{n}^2-3 \mathrm{n}=(\mathrm{n}-2) \cdot 180^{\circ}\quad\text{.... (1)} \end{aligned}$

Now according to question

$\begin{aligned} & a+(n-1) 6^{\circ}=219^{\circ} \\ & \Rightarrow a=225^{\circ}-6 n^{\circ}\quad\text{.... (2)} \end{aligned}$

Putting value of a from equation (2) in (1)

We get

$\begin{aligned} & \left(225 \mathrm{n}-6 \mathrm{n}^2\right)+3 \mathrm{n}^2-3 \mathrm{n}=180 \mathrm{n}-360 \\ & \Rightarrow 2 \mathrm{n}^2-42 \mathrm{n}-360=0 \\ & \Rightarrow \mathrm{n} 2-14 \mathrm{n}-120=0 \\ & \mathrm{n}=20,-6 \text { (rejected) } \end{aligned}$

$\begin{aligned} &\begin{aligned} & S_{11}=\frac{11}{2}(2 a+10 d)=88 \\ & a+5 d=8 \\ & a=8-5 \times \frac{3}{2}=\frac{1}{2} \end{aligned}\\ &\text { Roots are }\\ &\begin{aligned} & \mathrm{T}_{10}=\mathrm{a}+9 \mathrm{~d}=\frac{1}{2}+9 \times \frac{3}{2}=14 \\ & \mathrm{~T}_{11}=\mathrm{a}+10 \mathrm{~d}=\frac{1}{2}+10 \times \frac{3}{2}=\frac{31}{2} \\ & \frac{\mathrm{p}}{3}=\mathrm{T}_{10}+\mathrm{T}_{11}=14+\frac{31}{2}=\frac{59}{2} \\ & \mathrm{p}=\frac{177}{2} \\ & \frac{\mathrm{q}}{3}=\mathrm{T}_{10} \times \mathrm{T}_{11}=7 \times 31=217 \\ & \mathrm{q}=651 \\ & \mathrm{q}-2 \mathrm{p} \\ & =651-177 \\ & =474 \end{aligned} \end{aligned}$

$\begin{aligned} & \frac{1}{\alpha+1}+\frac{1}{\alpha+2}+\ldots+\frac{1}{\alpha+2012}- \\ & \left(\frac{1}{2 \times 21}+\frac{1}{4 \times 3}+\ldots+\frac{1}{2024} \cdot \frac{1}{2023}\right)=\frac{1}{2024} \\ & \quad \sum_{r=1}^{1012} \frac{1}{2 r(2 r-1)}=\sum_{r=1}^{1012}\left(\frac{1}{2 r-1}-\frac{1}{2 r}\right) \\ & \quad=\left(1-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)+\ldots+\left(\frac{1}{2023}-\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\frac{1}{5}+\ldots .+\frac{1}{2023}\right) \\ & \quad-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\ldots+\frac{1}{2024}\right) \\ & \quad=\left(1+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{1012}\right) \end{aligned}$

$\begin{aligned} & =\left(1+\frac{1}{2}+\frac{1}{3}+\ldots+\frac{1}{2023}\right)-\frac{1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1011}\right) \\ & \frac{-1}{2}\left(1+\frac{1}{2}+\ldots+\frac{1}{1012}\right) \\ & =\frac{1}{1012}+\frac{1}{1013}+\ldots+\frac{1}{2023}-\frac{1}{2024} \\ & \Rightarrow \alpha+1012=2023 \\ & \Rightarrow \alpha=1011 \end{aligned}$

First term is each row form pattern

$\begin{aligned} & \Rightarrow T_n=a n^2+b n+c \\ & \Rightarrow T_1=a+b+c=2 \\ & \Rightarrow T_2=4 a+2 b+c=5 \\ & \Rightarrow T_3=9 a+3 b+c=11 \\ & \Rightarrow 3 a+b=3 \\ & 5 a+b=6 \\ & \Rightarrow 2 a=3 \Rightarrow a=\frac{3}{2}, \quad b=\frac{-3}{2} \Rightarrow c=2 \\ & \Rightarrow T_n=\frac{3}{2} n^2-\frac{3(n)}{2}+2 \Rightarrow \frac{3 n^2-3 n+4}{2} \\ & T_{10}=\frac{3 \times 100-3 \times 10+4}{2}=\frac{274}{2}=137 \end{aligned}$

Terms in $10^{\text {th }}$ row is 10 with 3 differences

$\begin{aligned} & \Rightarrow 137,140,143 \ldots \\ & \Rightarrow \quad S_{10}=\frac{10}{2}(2 \times 137+(10-1) \times 3) \\ & =5(274+27)=5 \times 301=1505 \end{aligned}$

To solve this problem, we need to determine in which row the number $5310$ appears when positive integers are arranged in rows such that the $k^\text{th}$ row contains exactly $k$ numbers.

Understanding the Pattern

First row ($k = 1$): Contains 1 number.

Second row ($k = 2$): Contains 2 numbers.

Third row ($k = 3$): Contains 3 numbers.

…

$k^\text{th}$ row: Contains $k$ numbers.

Total Numbers Up to the $k^\text{th}$ Row

The total number of integers up to the $k^\text{th}$ row is given by the sum of the first $k$ natural numbers:

$ S(k) = 1 + 2 + 3 + \dots + k = \frac{k(k+1)}{2} $

Finding the Row Containing 5310

We need to find the smallest integer $k$ such that $S(k-1) < 5310 \leq S(k)$. This means:

$ \frac{(k-1)k}{2} < 5310 \leq \frac{k(k+1)}{2} $

Step 1: Estimate $k$

Let's approximate $k$ by solving the inequality:

$ \frac{k(k+1)}{2} \geq 5310 $

Multiply both sides by 2:

$ k(k+1) \geq 10620 $

This is a quadratic inequality. We can approximate $k$ by taking the square root:

$ k \approx \sqrt{10620} \approx 103 $

Step 2: Calculate $S(k)$ for $k = 102$ and $k = 103$

For $k = 102$:

$ S(102) = \frac{102 \times 103}{2} = \frac{10506}{2} = 5253 $

For $k = 103$:

$ S(103) = \frac{103 \times 104}{2} = \frac{10712}{2} = 5356 $

Step 3: Determine the Correct Row

Since $S(102) = 5253 < 5310 \leq 5356 = S(103)$, the number $5310$ lies between $5254$ and $5356$, inclusive. Therefore, it is in the $103^\text{rd}$ row.

Conclusion

The number $5310$ will be in the $103^\text{rd}$ row.

$\begin{aligned} \alpha= & \sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r \\ & =4 \sum_{r=0}^n r^2{ }^n C_r+2 \sum_{r=0}^n r \cdot{ }^n C_r+\sum_{r=0}^n{ }^n C_r \\ & =4 n(n+1) 2^{n-2}+2 \cdot n \cdot 2^{n-1}+2^n \\ & =2^n(n(n+1)+n+1)=2^n(n+1)^2 \\ & \beta=\sum_{r=0}^n\left(\frac{{ }^n C_r}{r+1}\right)+\left(\frac{1}{n+1}\right) \\ & (1+x)^n=\sum_{r=0}^n{ }^n C_r x^r \end{aligned}$

$\begin{aligned} & \int_\limits0^1(1+x)^n d x=\left.\sum_{r=0}^n \frac{{ }^n C_r x^{r+1}}{r+1}\right|_0 ^1=\sum_\limits{r=0} \frac{{ }^n C}{r+1} \\ & \left.\frac{(1+x)^{+1}}{n+1}\right|_0 ^1=\frac{2^n-1}{n+1} \\ \Rightarrow & \beta=\frac{2^{n+1}-1+1}{(n+1)}=\frac{2}{n+1} \\ \Rightarrow & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n \quad 1)}{\left(\frac{2^{n+1}}{n+1}\right)}=(n+1)^3 \in(140,281) \\ \Rightarrow & (n+1)^3=216 \\ \Rightarrow & n+1=6 \Rightarrow n=5 \end{aligned}$

$\begin{aligned} & (-x)(S(x))=\frac{(1+x)\left[(1+x)^{60}-1\right]}{(1+x-1)}-60(1+x)^{61} \\ & (-x) S(x)=\frac{(1+x)\left[(1+x)^{60}-1\right]}{x}-60(1+x)^{61} \\ & x S(x)=60(1+x)^{61}-\frac{(1+x)\left[(1+x)^{60}-1\right]}{x} \end{aligned}$

Multiplying $x$ on both side,

$x^2 S(x)=60 x(1+x)^{61}-(1+x)\left[(1+x)^{60}-1\right]$

Putting $x=60$

$\begin{aligned} & (60)^2 S(60)=60 \times 60(61)^{61}-(61)\left[61^{60}-1\right] \\ & =60 \times 60(61)^{61}-(61) \cdot 61^{60}+61 \\ & =(61)^{61}[60 \times 60-1]+61 \\ & =(3600-1) \cdot 61^{61}+61 \\ & a=3600-1, \quad b=61 \Rightarrow a+b=3660 \end{aligned}$

$\begin{aligned} & T_r=3 T_{r-1}+6^r \\ & \Rightarrow \text { solving homogenous part } \\ & T_r=3 T_{r-1} \\ & \Rightarrow x=3 \text { is the root } \end{aligned}$

$\therefore T_r=a .3^r$

Solving for particular part

$\begin{aligned} & T_r=b .6^r \\ & b .6^r=3 b 6^{r-1}+6^r \\ & \Rightarrow 6 b=3 b+6 \\ & \Rightarrow 3 b=6 \\ & \Rightarrow b=2 \\ & T_r=a^n+a^p \\ & T_r=a 3^{b r}+2.6^r \quad \text{.... (i)} \\ & T_r=3 T_{r-1}+6^r \end{aligned}$

Putting $r=2$

$T_2=18+36=54 \quad \text{.... (ii)}$

Using equation (i) and (ii)

$\begin{aligned} & 54=9 a+72 \Rightarrow-18=9 a \Rightarrow a=-2 \\ & \therefore T_r=2 \cdot 6^r-2 \cdot 3^r=2\left(6^r-3^r\right) \\ & \sum_{r=1}^n T_r=2 \sum 6^r-2 \sum 3^r \\ & =2 \cdot 6 \frac{\left(6^n-1\right)}{5}-2 \cdot 3 \frac{\left(3^n-1\right)}{2} \\ & =\frac{3}{5}\left(4 \cdot 6^n-5 \cdot 3^n+1\right) \\ & \therefore n^2-12 n+39=3 \\ & n^2-12 n+36=0 \\ & (n-6)^2=0 \\ & \therefore n=6 \end{aligned}$

$\begin{aligned} & S=1+\frac{\sqrt{3}-\sqrt{2}}{2 \sqrt{3}}+\frac{5-2 \sqrt{6}}{18}+\frac{9 \sqrt{3}-11 \sqrt{2}}{36 \sqrt{3}}+\ldots \infty \\\\ & =1+\frac{(1-\sqrt{2} / \sqrt{3})}{2}+\frac{(1-\sqrt{2} / \sqrt{3})^2}{6}+\frac{(1-\sqrt{2} / \sqrt{3})^3}{12}+\ldots \infty \end{aligned}$

$\text { let } 1-\frac{\sqrt{2}}{\sqrt{3}}=a$

$\begin{aligned} & S=1+\frac{a}{2}+\frac{a^2}{6}+\frac{a^3}{12}+\ldots \\\\ & =1+\left(1-\frac{1}{2}\right) a+\left(\frac{1}{2}-\frac{1}{3}\right) a^2+\left(\frac{1}{3}-\frac{1}{4}\right) a^3+\ldots \\\\ & =1+\left(a+\frac{a^2}{2}+\frac{a^3}{3} \ldots \infty\right)+\frac{1}{a}\left(\frac{-a^2}{2}-\frac{a^3}{3}-\frac{a^4}{4} \ldots \infty\right) \\\\ & =-\ln (1-a)+\frac{1}{a}\left(-a-\frac{a^2}{2}-\frac{a^3}{3} \ldots \infty\right)+2 \\\\ & =-\ln (1-a)+\frac{1}{a} \ln (1-a)+2 \\\\ & =2+\left(\frac{1}{a}-1\right) \ln (1-a) \\\\ & =2+\left(\frac{\sqrt{3}}{\sqrt{3}-\sqrt{2}}-1\right) \ln \left(1-1+\sqrt{\frac{2}{3}}\right) \\\\ & =2+\frac{\sqrt{2}}{\sqrt{3}}-\sqrt{2} \ln \sqrt{\frac{2}{3}} \\\\ & =2+\left(\frac{\sqrt{6}+2}{1} \cdot \frac{1}{2} \ln \frac{2}{3}\right) \\\\ & \therefore 2+\left(\sqrt{\frac{3}{2}}+1\right) \ln \frac{2}{3} \\\\ & \therefore 11 a+18 b=76 \end{aligned}$

Let $a_n=a+(n-1) d \forall n \in N$

$\begin{aligned} A_k & =\left(a_1^2-a_2^2\right)+\left(a_3^2-a_4^2\right)+\ldots a_{2 k-1}^2-a_{2 k}^2 \\ & =(-d)\left(a_1+a_2+\ldots+a_{2 k}\right) \\ & A_k=(-d k)(2 a+(2 k-1) d) \\ \Rightarrow & A_3=(-3 d)(2 a+5 d)=-153 \\ \Rightarrow & d(2 a+5 d)=51 \quad \text{... (i)}\\ & A_5=(-5 d)(2 a+9 d)=-435 \end{aligned}$

$\begin{aligned} \Rightarrow & d(2 a+9 d)=87 \\ \Rightarrow & 4 d^2=36 \Rightarrow d= \pm 3(d=3 \text { positive terms }) \\ \Rightarrow & 3(2 a+27)=87 \\ \Rightarrow & 2 a=29-27 \\ \Rightarrow & a=1 \\ & a_{17}-A_7=(a+16 d)-(-7 d)(2+13 d) \\ & =49+7 \times 3(2+39) \\ & =49+21 \times 41=910 \end{aligned}$

To solve this problem, let's first denote the three successive terms of a geometric progression (G.P.) with common ratio $r$ as $a$, $ar$, and $ar^2$, where $a$ is the first term and $r > 1$. These three terms represent the lengths of the sides of a triangle.

According to the triangle inequality theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side. Therefore, for the three terms to form a triangle, the following inequalities must hold:

$ 1) \ a + ar > ar^2 \\\\ $

$2) \ a + ar^2 > ar $

$ 3) \ ar + ar^2 > a $

Given that $r > 1$, inequalities 2 and 3 will always hold because:

$ ar < ar^2 \ \text{and} \ a < ar, $

indicating that both $a + ar^2$ and $ar + ar^2$ will be greater than $ar$ and $a$ respectively. Therefore, we only need to check the first inequality to ensure that the three terms can form a triangle:

$ a + ar > ar^2 $

Simplifying this, we get:

$ a(1 + r) > a r^2 $

Since $a$ is positive (as it represents the length of a side of a triangle), we can divide both sides of the inequality by $a$ without changing the direction of the inequality:

$ 1 + r > r^2 $

We can then subtract $r$ from both sides:

$ 1 > r^2 - r $

Simplifying the right side by factoring $r$:

$ 1 > r(r - 1) $

Given that $r > 1$, the quantity $(r - 1)$ is positive; hence, $r(r - 1)$ is also positive. This means the actual value for $r$ to satisfy the inequality is within the interval $(1, \sqrt{2})$ because $r(r - 1)$ increases with increasing $r$, and it would be 1 when $r = \sqrt{2}$. It should be greater than 1, and less than $\sqrt{2}$ such that $r^2 - r$ stays below 1.

Now let’s consider the expressions $[r]$ and $[-r]$. The symbol $[x]$ denotes the greatest integer less than or equal to $x$ (also known as the floor function).

Since $1 < r < \sqrt{2}$, $[r] = 1$, because 1 is the greatest integer less than $r$ within that interval.

For $[-r]$, we need the greatest integer less than or equal to $-r$. Since $-r$ is negative and less than $-1$ (because $r > 1$), $[-r] = -2$, as this is the greatest integer that does not exceed the negative value of $r$ (which lies between $-\sqrt{2}$ and $-1$).

Now we can substitute these values into the expression:

$ 3[r] + [-r] = 3 \cdot 1 + (-2) = 3 - 2 = 1 $

Therefore $3[r] + [-r]$ is equal to 1.

To find the common terms in the two given arithmetic progressions (AP), we need to first identify the common difference for each sequence and then find the sequence that represents their overlap by employing the concept of least common multiple (LCM).

The first AP is:

$3, 7, 11, 15, \ldots, 403$

The common difference ($d_1$) for the first AP can be calculated by subtracting the first term from the second term:

$d_1 = 7 - 3 = 4$

The second AP is:

$2, 5, 8, 11, \ldots, 404$

The common difference ($d_2$) for the second AP is:

$d_2 = 5 - 2 = 3$

To find the terms common to both sequences, we need to find a term that appears in both sequences. Any common term must be of the form $3 + 4k$ and $2 + 3l$ for some integers $k$ and $l$. We want to find when these two forms will give us the same number, so we set them equal to each other:

$3 + 4k = 2 + 3l$

Rearranging the terms gives us:

$4k - 3l = 2 - 3$

This simplifies to:

$4k - 3l = -1 ....... (1)$

The solutions to equation $(1)$ will give us the common terms. Notice this is a Diophantine equation (A Diophantine equation is a polynomial equation, usually with two or more variables,) and has an infinite number of solutions. Let's find one such solution. We can see that:

$k = 1 \quad \text{yields} \quad 4(1) - 3l = -1 \implies 4 - 3l = -1 \implies 3l = 5 \implies l = 1\frac{2}{3}$

This is not an integer solution for $l$, so $k = 1$ does not work. Trying $k = 2$ gives:

$4(2) - 3l = -1 \implies 8 - 3l = -1 \implies 3l = 9 \implies l = 3$

Now we've found integers $k = 2$ and $l = 3$ that satisfy the equation. The corresponding term in both sequences would be:

$3 + 4(2) = 3 + 8 = 11 \quad \text{and} \quad 2 + 3(3) = 2 + 9 = 11$

Since $11$ is a common term, we can assert that every common term in both APs will be of the form $11 + m(4 \times 3)$, where $m$ is a non-negative integer, and $4 \times 3 = 12$ is the LCM of the common differences of the two APs. Thus, the general form for the common terms would be:

$11 + 12m$

Now we are to find all terms that are common up to $403$ in the first sequence and up to $404$ in the second sequence. Because the first sequence doesn't exceed $403$, we'll use this as our limit:

$11 + 12m \leq 403$

To find the largest possible integer value for $m$, we solve the inequality:

$12m \leq 403 - 11$

$12m \leq 392$

$m \leq 32\frac{2}{3}$

Since $m$ has to be an integer, the largest possible value for $m$ is $32$. Therefore, the common terms are generated by $m = 0, 1, 2, \ldots, 32$. There are $32 + 1 = 33$ terms in total.

We will now sum these up. The sum of an AP is given by the formula:

$S = \frac{n}{2}(a_1 + a_n)$

Where $S$ is the sum, $n$ is the number of terms, $a_1$ is the first term, and $a_n$ is the last term. Using the formula:

$S = \frac{33}{2}(11 + (11 + 12 \times 32))$

$S = \frac{33}{2}(11 + 11 + 384)$

$S = \frac{33}{2}(11 + 11 + 384)$

$S = \frac{33}{2}(406)$

$S = 33 \times 203$

$S = 6699$

Therefore, the sum of the common terms in the two arithmetic progressions is 6699.

$\begin{aligned} & \mathrm{S}_{\mathrm{n}}= 3+7+11+\ldots \ldots \mathrm{n} \text { terms } \\ &=\frac{\mathrm{n}}{2}(6+(\mathrm{n}-1) 4)=3 \mathrm{n}+2 \mathrm{n}^2-2 \mathrm{n} \\ &=2 \mathrm{n}^2+\mathrm{n} \\ & \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{S}_{\mathrm{k}}=2 \sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K}^2+\sum_{\mathrm{k}=1}^{\mathrm{n}} \mathrm{K} \\ &=2 \cdot \frac{\mathrm{n}(\mathrm{n}+1)(2 \mathrm{n}+1)}{6}+\frac{\mathrm{n}(\mathrm{n}+1)}{2} \\ &=\mathrm{n}(\mathrm{n}+1)\left[\frac{2 \mathrm{n}+1}{3}+\frac{1}{2}\right] \\ &=\frac{\mathrm{n}(\mathrm{n}+1)(4 \mathrm{n}+5)}{6} \\ & \Rightarrow 40<\frac{6}{\mathrm{n}(\mathrm{n}+1)} \sum_{k=1}^n S_{\mathrm{k}}<42 \\ & 40<4 \mathrm{n}+5<42 \\ & 35<4 n<37 \\ & \mathrm{n}=9 \end{aligned}$

$\begin{gathered} \alpha=1^2+4^2+8^2 \ldots . \\ t_n=a^2+b n+c \end{gathered}$

$\begin{aligned} & 1=a+b+c \\ & 4=4 a+2 b+c \\ & 8=9 a+3 b+c \end{aligned}$

On solving we get, $\mathrm{a}=\frac{1}{2}, \mathrm{~b}=\frac{3}{2}, \mathrm{c}=-1$

$\begin{aligned} & \alpha=\sum_{n=1}^{10}\left(\frac{n^2}{2}+\frac{3 n}{2}-1\right)^2 \\ & 4 \alpha=\sum_{n=1}^{10}\left(n^2+3 n-2\right)^2, \beta=\sum_{n=1}^{10} n^4 \\ & 4 \alpha-\beta=\sum_{n=1}^{10}\left(6 n^3+5 n^2-12 n+4\right)=55(353)+40 \end{aligned}$

$8=\frac{3}{1-\frac{1}{4}}+\frac{p \cdot \frac{1}{4}}{\left(1-\frac{1}{4}\right)^2}$

$\text { (sum of infinite terms of A.G.P }=\frac{a}{1-r}+\frac{d r}{(1-r)^2} \text { ) }$

$\Rightarrow \frac{4 p}{9}=4 \Rightarrow p=9$

We can rewrite the given series as follows :

$S = \left(\frac{1}{2}-\frac{1}{3}\right) + \left(\frac{1}{4}-\frac{1}{6}+\frac{1}{9}\right) + \left(\frac{1}{8}-\frac{1}{12}+\frac{1}{18}-\frac{1}{27}\right) + \left(\frac{1}{16}-\frac{1}{24}+\frac{1}{36}-\frac{1}{54}+\frac{1}{81}\right) + \ldots$

The first few terms of the series are :

$S = \left(\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \frac{1}{16} + \ldots\right) - \left(\frac{1}{3} + \frac{1}{6} + \frac{1}{12} + \frac{1}{24} + \ldots\right) + \left(\frac{1}{9} + \frac{1}{18} + \frac{1}{36} + \ldots\right) - \left(\frac{1}{27} + \frac{1}{54} + \ldots\right) + \ldots$

We can now see that each group of terms forms a geometric series with a common ratio of $\frac{1}{2}$:

$S = \frac{\frac{1}{2}}{1-\frac{1}{2}} - \frac{\frac{1}{3}}{1-\frac{1}{2}} + \frac{\frac{1}{9}}{1-\frac{1}{2}} - \frac{\frac{1}{27}}{1-\frac{1}{2}} + \ldots$

The series can be rewritten as :

$S = 2 \left(\frac{1}{{2}} - \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)$

Now, we can simplify and rewrite the series inside the parentheses as:

$S = 2 \left[\frac{1}{{2}} + (- \frac{1}{3} + \frac{1}{9} - \frac{1}{27} + \ldots\right)]$

The series inside the parentheses is an infinite geometric series with the first term $a = - \frac{1}{3}$ and the common ratio $r = -\frac{1}{3}$:

$S = 2 \left(\frac{1}{{2}} +\frac{a}{1-r}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{1+\frac{1}{3}}\right) = 2 \left(\frac{1}{{2}} + \frac{- \frac{1}{3}}{\frac{4}{3}}\right) $

= $2\left( {{1 \over 2} - {1 \over 3} \times {3 \over 4}} \right)$

$ = 2\left( {{1 \over 2} - {1 \over 4}} \right) = 2\left( {{1 \over 4}} \right) = {1 \over 2}$

Thus, the sum of the series is $\frac{1}{2}$, and $\alpha = 1$ and $\beta = 2$ are co-prime.

Therefore, $\alpha + 3\beta = 1 + 6 = 7$

$ \begin{aligned} & \sum_{r=1}^{10}\left(2 \cdot(2 r)^2-(2 r+1)^2\right) \\\\ & =\sum_{r=1}^{10}\left(8 r^2-4 r^2-4 r-1\right) \\\\ & =\sum_{r=1}^{10}\left(4 r^2-4 r-1\right) \\\\ & =\frac{4 \cdot 10 \cdot 11 \cdot 21}{6}-4 \frac{10 \cdot 11}{2}-10 \\\\ & =44 \cdot 35-220-10 \\\\ & =1540-230=1310 \end{aligned} $

From the given series :

$10 = 1+\frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$

We isolate the 1 to get :

$9 = \frac{4}{k}+\frac{8}{k^2}+\frac{13}{k^3}+\frac{19}{k^4}+\ldots$ .......(1)

Divide each term in the equation by $k$ :

$\frac{9}{k} = \frac{4}{k^2}+\frac{8}{k^3}+\frac{13}{k^4}+\ldots$ .......(2)

Subtracting the second equation from the first, we obtain :

$9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$

This is equivalent to :

$S = 9\left(1-\frac{1}{k}\right) = \frac{4}{k} + \frac{4}{k^2} + \frac{5}{k^3} + \frac{6}{k^4} + \ldots$ .........(3)

Divide both sides by $k$ again, we get : $\frac{S}{k} = \frac{4}{k^2} + \frac{4}{k^3} + \frac{5}{k^4} + \ldots$ ..........(4)

Subtracting equation (4) from (3), we have :

$(1-\frac{1}{k})S = \frac{4}{k} + \frac{1}{k^3} + \frac{1}{k^4} + \ldots$

In this equation, you've treated the infinite series on the right-hand side as a geometric series with a ratio of $1/k$, so the sum of this series can be expressed as $\frac{1/k^3}{1 - 1/k}$.

Therefore, we get :

$9\left(1-\frac{1}{k}\right)^2 = \frac{4}{k} + \frac{1/k^3}{1 - 1/k}.$

Now, simplifying this equation, we get :

$9(k-1)^2 = 4k^2 - 4k + 1$

$ \Rightarrow $$9(k^2 - 2k + 1) = 4k^2 - 4k + 1$

$ \Rightarrow $$9k^2 - 18k + 9 = 4k^2 - 4k + 1$

$ \Rightarrow $$9k^2 - 4k^2 - 18k + 4k + 9 - 1 = 0$

$ \Rightarrow $$5k^2 - 14k + 8 = 0$

This is a standard form quadratic equation, $ax^2 + bx + c = 0$, where $a=5$, $b=-14$, and $c=8$.

The solutions can be found using the quadratic formula, $k = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Substituting the values for $a$, $b$, and $c$, we have :

$ \Rightarrow $$k = \frac{14 \pm \sqrt{(-14)^2 - 4\times5\times8}}{2\times5}$

$ \Rightarrow $$k = \frac{14 \pm \sqrt{196 - 160}}{10}$

$ \Rightarrow $$k = \frac{14 \pm \sqrt{36}}{10}$

$ \Rightarrow $$k = \frac{14 \pm 6}{10}$

So, the solutions are $k = 2$ or $k = 0.8$. However, since the question mentions $k \in \mathbb{N}$, i.e., $k$ is a natural number, the only valid solution is $k = 2$.

We have, $S=109+\frac{108}{5}+\frac{107}{5^2}+\ldots+\frac{2}{5^{107}}+\frac{1}{5^{108}}$ ...........(i)

$\frac{S}{5}=\frac{109}{5}+\frac{108}{5^2}+\frac{107}{5^3}+\ldots .+\frac{2}{5^{108}}+\frac{1}{5^{109}}$ .............(ii)

On subtracting Eq. (ii) from Eq. (i), we get

$ \begin{aligned} \frac{4 S}{5} & =109-\frac{1}{5}-\frac{1}{5^2}-\ldots .-\frac{1}{5^{108}}-\frac{1}{5^{109}} \\\\ \frac{4 S}{5} & =109-\left[\frac{1}{5}+\frac{1}{5^2}+\ldots+\frac{1}{5^{108}}+\frac{1}{5^{109}}\right] \end{aligned} $

This form a GP with $r=\frac{1}{5}$

$ \begin{aligned} \frac{4 S}{5} & =109-\frac{1}{5}\left[\frac{1-\frac{1}{5^{109}}}{1-\frac{1}{5}}\right]\\\\ &=109-\frac{1}{4}\left[1-\frac{1}{5^{109}}\right] \\\\ & =109-\frac{1}{4}+\frac{1}{4} \times \frac{1}{5^{109}} \end{aligned} $

$ \therefore $ $ 4 S=\frac{5}{4}\left[435+\frac{1}{5^{109}}\right] $

$ \Rightarrow 16 S=2175+\frac{1}{5^{108}} $

$ 16 S-(25)^{-54}=2175 $

Since, common ratio of A.G.P. is 2 therefore A.G.P. can be taken as

$ \begin{aligned} & \frac{(a-2 d)}{4}, \frac{(c-d)}{2}, a, 2(a+d), 4(a+2 d) \\\\ & \text { or } a_1, a_2, 2, a_3, a_4 \text { (Given) } \\\\ & \Rightarrow a=2 \end{aligned} $

also sum of thes A.G.P. is $\frac{49}{2}$

$ \begin{aligned} & \Rightarrow \frac{2-2 d}{4}+\frac{2-d}{2}+2+2(2+d)+4(2+2 d)=\frac{49}{2} \\\\ & \Rightarrow \frac{1}{4}[2-2 d+4-2 d+8+16+8 d+32+32 d]=\frac{49}{2} \\\\ & \Rightarrow 36 d+62=98 \end{aligned} $

$ \Rightarrow 36 d=36 \Rightarrow d=1 $

Hence, $a_4=4(a+2 d)=4(2+2 \times 1)=16$

The sum of all those terms, of the arithmetic progression $3,8,13 \ldots, 373$.

Which are not divisible by 3 is

$ \begin{aligned} & =(3+8+13+18+\ldots+373) -(3+18+33+\ldots+363) \\\\ & =\frac{75}{2}(3+373)-\frac{25}{2}(3+363) \\\\ & =\frac{75}{2} \times 376-\frac{25}{2} \times 366 \\\\ & =75 \times 188-25 \times 183 \\\\ & =14100-4575 \\\\ & =9525 \end{aligned} $

$\because \frac{1}{x}, \frac{1}{y}, \frac{1}{z}$ are in A.P.

$ \Rightarrow \frac{1}{x}+\frac{1}{z}=\frac{2}{y} $ ........... (i)

and $x, \sqrt{2} y, z$ are in G.P.

$ \Rightarrow 2 y^2=x z $ .......... (ii)

from (i), $\frac{2}{y}=\frac{x+z}{x z}=\frac{x+z}{2 y^2}$

$ \Rightarrow 4 y=x+z $

$ \begin{aligned} & \text { Also, } x y+y z+z x=\frac{3}{\sqrt{2}} x y z \\\\ & y(4 y)+x z=\frac{3}{\sqrt{2}}\left(2 y^2\right) y \\\\ & \Rightarrow 4 y^2+2 y^2=3 \sqrt{2} y^3 \\\\ & \Rightarrow 6 y^2=3 \sqrt{2} y^3 \Rightarrow y=\sqrt{2} \\\\ & \therefore 3(x+y+z)^2=3(5 y)^2=3(5 \sqrt{2})^2 \\\\ & =150 \end{aligned} $

$\begin{aligned} &(20)^{19}+2(21)(20)^{18}+3(21)^2(20)^{17} \\ & \quad+\ldots \ldots+20(21)^{19}=k(20)^{19} \\\\ & \Rightarrow(20)^{19}\left[1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19}\right]=k(20)^{19} \\\\ & \Rightarrow k=1+2\left(\frac{21}{20}\right)+3\left(\frac{21}{20}\right)^2+\ldots+20\left(\frac{21}{20}\right)^{19} ..........(i)\end{aligned}$

Now,

$\begin{aligned} k\left(\frac{21}{20}\right)=\left(\frac{21}{20}\right) & +2\left(\frac{21}{20}\right)^2+3\left(\frac{21}{20}\right)^3 +\ldots+20\left(\frac{21}{20}\right)^{20}\end{aligned}$ ........(ii)

On subtracting Equation (ii) from Equation (i), we get

$ \begin{aligned} & k\left(\frac{-1}{20}\right)=1+\frac{21}{20}+\left(\frac{21}{20}\right)^2+\ldots \ldots+\left(\frac{21}{20}\right)^{19}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=\frac{1\left(\left(\frac{21}{20}\right)^{20}-1\right)}{\frac{21}{20}-1}-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=20\left(\left(\frac{21}{20}\right)^{20}-1\right)-20\left(\frac{21}{20}\right)^{20} \\\\ & =20\left(\frac{21}{20}\right)^{20}-20-20\left(\frac{21}{20}\right)^{20} \\\\ & \Rightarrow k\left(\frac{-1}{20}\right)=-20 \\\\ & \Rightarrow k=400 \end{aligned} $

$ \begin{array}{ll} 3,7,11,15, \ldots \ldots \ldots . .399 : & \mathrm{~d}_1=4 \\\\ 2,5,8,11, \ldots \ldots \ldots \ldots, 359 : & \mathrm{~d}_2=3 \\\\ 2,7,12,17, \ldots \ldots, 197 : & \mathrm{~d}_3=5 \\\\ \operatorname{LCM}\left(\mathrm{d}_1, \mathrm{~d}_2, \mathrm{~d}_3\right)=60 & \end{array} $

Common terms are 47, 107, 167

Sum $=321$

$ \begin{aligned} & a_1+a_2+a_3+a_4=50 \\\\ & \Rightarrow 32+6 d=50 \\\\ & \Rightarrow d=3 \\\\ & \text { and, } a_{n-3}+a_{n-2}+a_{n-1}+a_n=170 \\\\ & \Rightarrow 32+(4 n-10) \cdot 3=170 \\\\ & \Rightarrow \mathrm{n}=14 \\\\ & a_7=26, a_8=29 \\\\ & \Rightarrow a_7 \cdot a_8=754 \end{aligned} $

$S=1^{2}-2.3^{2}+3.5^{2}-4.7^{2}+\ldots \ldots+15.29^{2}$

Separating odd placed and even placed terms we get

$ \begin{aligned} & \mathrm{S}=\left(1.1^2+3.5^2+\ldots .15 .(29)^2\right)-\left(2.3^2+4.7^2\right. \\ & +\ldots .+14 .(27)^2 \end{aligned} $

$ \begin{aligned} & =\sum_{r=1}^{8}(2 r-1)(4 r-3)^{2}-\sum_{r=1}^{7} 2 r(4 r-1)^{2} \\\\ & =\sum_{r=1}^{8} (32 r^{3}-64 r^{2}+42 r-9)-2\sum_{r=1}^{7} 16 r^{3}-8 r^{2}+r \\\\ & =32 \times 36^{2}-64 \times 204+1512-72 \\\\ & -2\left(16 \times 28^{2}-1120+28\right) \\\\ & =6592 \end{aligned} $

$a_{11}=18$

$ \begin{aligned} & a+10 d=18 \\\\ & a_{5}=2 a_{7} \\\\ & a+4 d=2(a+6 d) \\\\ & a=-8 d \end{aligned} $

(i) and (ii) $\Rightarrow a=-72, d=9$.

On rationalising the denominator, given expression

$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{11}}}{-d}+\frac{\sqrt{a_{11}}-\sqrt{a_{12}}}{-d}+\ldots+\frac{\sqrt{a_{17}}-\sqrt{a_{18}}}{-d}\right]$

$=12\left[\frac{\sqrt{a_{10}}-\sqrt{a_{18}}}{-d}\right]$

$=12\left[\frac{\sqrt{a_{11}-d}-\sqrt{a_{11}+7 d}}{-d}\right]$

$=12\left[\frac{\sqrt{18-9}-\sqrt{18+63}}{-9}\right]$ $=12 \times \frac{2}{3}=8$

First common term is 11

Common difference of series of common terms is LCM (4, 5) = 20

$a_8=a+7d$

$=11+7\times20=151$

$\sum\limits_{n = 0}^\infty {{{{n^3}(2n!) + (2n - 1)(n!)} \over {n!\,.\,(2n)!}}} $

$ = \sum\limits_{n = 0}^\infty {{{{n^3}} \over {n!}} + {{2n - 1} \over {2n!}}} $

$ = \sum\limits_{n = 0}^\infty {{3 \over {(n - 2)!}} + {1 \over {(n - 3)!}} + {1 \over {(n - 1)!}} + {1 \over {(2n - 1)!}} - {1 \over {(2n)!}}} $

$ = 3e + e + e - {1 \over e}$

$ = 5e - {1 \over e}$

$\therefore$ $a = 5,b = - 1,c = 0$

$\therefore$ ${a^2} - b + c = 26$

$\because$ ${a_n} = {a_{n - 1}} + (n - 1)$ and ${a_1} = {b_1} = 1$

${b_n} = {b_{n - 1}} + {a_{n - 1}}$

$\therefore$ ${b_{n + 1}} = 2{b_n} - {b_{n - 1}} + n - 1$

$n$	$b_n$	$b_n-n$
1	1	0
2	2	0
3	4	1
4	8	4
5	15	10
6	26	20
7	42	35
8	64	56
9	93	84
10	130	120

$\therefore$ $2S - T = \left( {\sum\limits_{n = 1}^8 {{{{b_n} - n} \over {{2^{n - 1}}}}} } \right) + {{{b_9}} \over {{2^8}}} + {{{b_{10}}} \over {{2^9}}}$

$ = {{461} \over {128}}$

$\therefore$ ${2^7}(2S - T) = 461$

$\{ {a_k}\} $ be a G.P. with ${a_1} = 4,r = {r_1}$

And

$\{ {b_k}\} $ be G.P. with ${b_1} = 4,r = {r_2}$ $({r_1} < {r_2})$

Now

${C_k} = {a_k} + {b_k}$

${c_1} = 4 + 4 = 8$ and ${c_2} = 5$

${a_2} + {b_2} = 5$

$\therefore$ ${r_1} + {r_2} = {5 \over 4}$

and ${c_3} = {{13} \over 4} \Rightarrow r_4^2 + r_2^2 = {{13} \over {16}}$

$\therefore$ ${{25} \over {16}} - 2{r_1}{r_2} = {{13} \over {16}} \Rightarrow 2{r_1}{r_2} = {3 \over 4}$

$\therefore$ ${r_2} - {r_1} = \sqrt {{{25} \over {16}} - {3 \over 2}} = {1 \over 4}$

$\therefore$ ${r_2} = {3 \over 4},{r_1} = {1 \over 2}$

$\therefore$ ${a_6} = 4 \times {1 \over {{2^5}}} = {1 \over 8},{b_4} = 4 \times {{27} \over {64}} = {{27} \over {16}}$

and $\sum\limits_{K = 1}^\infty {{C_K} = 4\left[ {{1 \over {1 - {1 \over 2}}} + {1 \over {1 - {3 \over 4}}}} \right] = 24} $

$\therefore$ $\sum\limits_{K = 1}^\infty {{C_K} - (12{a_6} + 8{b_4}) = 09} $

Let $r$ be the common ratio of the G.P

$\therefore a_{1} r^{3} \times a_{1} r^{5}=9$

$a_{1}^{2} r^{8}=9 \Rightarrow a_{1} r^{4}=3$

And

$ \begin{aligned} & a_{1}\left(r^{4}+r^{6}\right)=24 \\\\ \Rightarrow & 3\left(1+r^{2}\right)=24 \\\\ \therefore & r^{2}=7 \text { and } a_{1}=\frac{3}{49} \end{aligned} $

Now

$ \begin{aligned} & a_{1} a_{9}+a_{2} a_{4} a_{9}+a_{5}+a_{7} \\\\ & =a_{1}^{2} r^{8}+a_{1}^{3} r^{12}+24 \\\\ & =24+\frac{9}{7^{4}} \times 7^{4}+\frac{27}{7^{6}} \cdot 7^{6}=60 \end{aligned} $

$ \begin{aligned} & \mathrm{a}, \mathrm{b}, \frac{1}{18} \rightarrow \mathrm{GP} \\\\ & \frac{\mathrm{a}}{18}=\mathrm{b}^2\quad...(i) \\\\ & \frac{1}{\mathrm{a}}, 10, \frac{1}{\mathrm{~b}} \rightarrow \mathrm{AP} \\\\ & \frac{1}{\mathrm{a}}+\frac{1}{\mathrm{~b}}=20 \\\\ & \Rightarrow \mathrm{a}+\mathrm{b}=20 \mathrm{ab}, \text { from eq. (i) } ; \text { we get } \\\\ & \Rightarrow 18 \mathrm{~b}^2+\mathrm{b}=360 \mathrm{~b}^3 \\\\ & \Rightarrow 360 \mathrm{~b}^2-18 \mathrm{~b}-1=0 \quad\{\because \mathrm{b} \neq 0\} \\\\ & \Rightarrow \mathrm{b}=\frac{18 \pm \sqrt{324+1440}}{720} \\\\ & \Rightarrow b =\frac{18+\sqrt{1764}}{720} \quad\{\because \mathrm{b}>0\} \\\\ & \Rightarrow b = \frac{1}{12} \\\\ & \Rightarrow \mathrm{a}=18 \times \frac{1}{144}=\frac{1}{8} \\\\ & \text{Now},16\mathrm{a}+12 \mathrm{~b}=16 \times \frac{1}{8}+12 \times \frac{1}{12}=3 \end{aligned} $

Given $\frac{1^{3}+2^{3}+3^{3}+\ldots \text { up to } n \text { terms }}{1.3+2.5+3.7+\ldots \text { up to } n \text { terms }}=\frac{9}{5}$

Now

Let $S=1.3+2.5+3.7+\ldots$

$ \begin{aligned} & T_{n}=n \cdot(2 n+1) \\\\ & \therefore S=\frac{2 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2} \\\\ & \Rightarrow \frac{\left(\frac{n(n+1)}{2}\right)^{2}}{n(n+1)\left[\frac{2 n+1}{3}+\frac{1}{2}\right]}=\frac{9}{5} \\\\ & \Rightarrow 5 n^{2}-19 n-30=0 \\\\ & \Rightarrow(5 n+6)(n-5)=0 \\\\ & \therefore n=5 \end{aligned} $

$T_{4}=500$

$ a r^{3}=500 \Rightarrow a=\frac{500}{r^{3}} $

Now,

$ \begin{aligned} & S_{6} > S_{5}+1 \\\\ & \frac{a\left(1-r^{6}\right)}{1-r}-\frac{a\left(1-r^{5}\right)}{1-r} > 1 \\\\ & a r^{5} > 1 \\\\ & \text { Now, } r=\frac{1}{m} \text { and } a=\frac{500}{r^{3}} \\\\ & \Rightarrow \quad m^{2} < 500 \\\\ & \because m > 0 \Rightarrow m \in(0,10 \sqrt{5}) \\\\ & \quad S_{7} < S_{6}+\frac{1}{2} \\\\ & \quad \frac{a\left(1-r^{6}\right)}{1-r}<\frac{a\left(1-r^{6}\right)}{1-r}+\frac{1}{2} \\\\ & \quad a r^{6}<\frac{1}{2} \end{aligned} $

$\because r=\frac{1}{m}$ and $a=\frac{500}{r^{5}}$

$ \frac{1}{m^{3}}<\frac{1}{1000} $

$\Rightarrow m \in(10, \infty)$

Possible values of $m$ is $\{11,12,....22 \}$

$\because m \in N$

Total 12 values

Given

$S = {{{a_1}} \over 2} + {{{a_2}} \over {{2^2}}} + {{{a_3}} \over {{2^3}}} + {{{a_4}} \over {{2^4}}}\, + \,.....\,\infty $

${{{1 \over 2}S = {{{a_1}} \over {{2^2}}} + {{{a_2}} \over {{2^3}}}\, + \,.........\,\infty } \over {{S \over 2} = {{{a_1}} \over 2} + {{({a_2} + {a_1})} \over {{2^2}}} + {{({a_3} + {a_2})} \over {{2^3}}}\, + \,......\,\infty }}$

$ \Rightarrow {S \over 2} = {{{a_1}} \over 2} + {d \over 2}$

$ \Rightarrow {a_1} + d = {a_2} = 4 \Rightarrow 4{a_2} = 16$

$S = {1 \over {2 \times 3 \times 4}} + {1 \over {3 \times 4 \times 5}} + {1 \over {4 \times 5 \times 6}}\, + \,....\, + \,{1 \over {100 \times 101 \times 102}}$

$ = {1 \over {(3 - 1)\,.\,1}}\left[ {{1 \over {2 \times 3}} - {1 \over {101 \times 102}}} \right]$

$ = {1 \over 2}\left( {{1 \over 6} - {1 \over {101 \times 102}}} \right)$

$ = {{143} \over {102 \times 101}} = {k \over {101}}$

$\therefore$ $34k = 286$

${1 \over {{3^{12}}}} + 5\left( {{{{2^0}} \over {{3^{12}}}} + {{{2^1}} \over {{3^{11}}}} + {{{2^2}} \over {{3^{10}}}}\, + \,.......\, + \,{{{2^{11}}} \over 3}} \right) = {2^n}\,.\,m$

$ \Rightarrow {1 \over {{3^{12}}}} + 5\left( {{1 \over {{3^{12}}}}{{\left( {{{(6)}^2} - 1} \right)} \over {(6 - 1)}}} \right) = {2^n}\,.\,m$

$ \Rightarrow {1 \over {{3^{12}}}} + {5 \over 5}\left( {{1 \over {{3^{12}}}}\,.\,{2^{12}}\,.\,{3^{12}} - {1 \over {{3^{12}}}}} \right) = {2^n}\,.\,m$

$ \Rightarrow {1 \over {{3^{12}}}} + {2^{12}} - {1 \over {{3^{12}}}} = {2^n}\,.\,m$

$ \Rightarrow {2^n}\,.\,m = {2^{12}}$

$ \Rightarrow m = 1$ and $n = 12$

$m\,.\,n = 12$

${T_n} = {{\sum\limits_{k = 1}^n {\left[ {{{(2k)}^3} - {{(2k - 1)}^3}} \right]} } \over {n(4n + 3)}}$

$ = {{\sum\limits_{k = 1}^n {4{k^2} + {{(2k - 1)}^2} + 2k(2k - 1)} } \over {n(4n + 3)}}$

$ = {{\sum\limits_{k = 1}^n {(12{k^2} - 6k + 1)} } \over {n(4n + 3)}}$

$ = {{2n(2{n^2} + 3n + 1) - 3{n^2} - 3n + n} \over {n(4n + 3)}}$

$ = {{{n^2}(4n + 3)} \over {n(4n + 3)}} = n$

$\therefore$ ${T_n} = n$

${S_n} = \sum\limits_{n = 1}^{15} {{T_n} = {{15 \times 16} \over 2} = 120} $

$\sum\limits_{k = 1}^{10} {{k \over {{k^4} + {k^2} + 1}}} $

$ = {1 \over 2}\left[ {\sum\limits_{k = 1}^{10} {\left( {{1 \over {{k^2} - k + 1}} - {1 \over {{k^2} + k + 1}}} \right)} } \right.$

$ = {1 \over 2}\left[ {1 - {1 \over 3} + {1 \over 3} - {1 \over 7} + {1 \over 7} - {1 \over {13}}\, + \,...\, + \,{1 \over {91}} - {1 \over {111}}} \right]$

$ = {1 \over 2}\left[ {1 - {1 \over {111}}} \right] = {{110} \over {2\,.\,111}} = {{55} \over {111}} = {m \over n}$

$\therefore$ $m + n = 55 + 111 = 166$

${d_1} = {{199 - 100} \over 2} \notin I$

${d_2} = {{199 - 100} \over 3} = 33$

${d_3} = {{199 - 100} \over 4} \notin I$

${d_n} = {{199 - 100} \over {i + 1}} \in I$

${d_i} = 33 + 11,\,9$

Sum of CD's $ = 33 + 11 + 9$

$ = 53$

Given series

$\therefore$ 11^th set will have $1 + (10)2 = 21$ term

Also upto 10^th set total $3 \times k$ type terms will be $1 + 3 + 5\, + \,......\, + \,19 = 100 - $ term

$\therefore$ Set $11 = \{ 3 \times 101,\,3 \times 102,\,......\,3 \times 121\} $

$\therefore$ Sum of elements $ = 3 \times (101 + 102\, + \,...\, + \,121)$

$ = {{3 \times 222 \times 21} \over 2} = 6993$

$\because$ Roots of $2 a x^{2}-8 a x+1=0$ are $\frac{1}{p}$ and $\frac{1}{r}$ and roots of $6 b x^{2}+12 b x+1=0$ are $\frac{1}{q}$ and $\frac{1}{s}$.

Let $\frac{1}{p}, \frac{1}{q}, \frac{1}{r}, \frac{1}{s}$ as $\alpha-3 \beta, \alpha-\beta, \alpha+\beta, \alpha+3 \beta$

So sum of roots $2 \alpha-2 \beta=4$ and $2 \alpha+2 \beta=-2$

Clearly $\alpha=\frac{1}{2}$ and $\beta=-\frac{3}{2}$

Now product of roots, $\frac{1}{p} \cdot \frac{1}{r}=\frac{1}{2 a}=-5 \Rightarrow \frac{1}{a}=-10$

and $\frac{1}{q} \cdot \frac{1}{x}=\frac{1}{6 b}=-8 \Rightarrow \frac{1}{b}=-48$

So, $\frac{1}{a}-\frac{1}{b}=38$

Given,

${a_n} = {a_{n - 1}} + 2$

$ \Rightarrow {a_n} - {a_{n - 1}} = 2$

$\therefore$ In this series between any two consecutives terms difference is 2. So this is an A.P. with common difference 2.

Also given ${a_1} = 1$

$\therefore$ Series is = 1, 3, 5, 7 ......

$\therefore$ ${a_n} = 1 + (n - 1)2 = 2n - 1$

Also ${b_n} = {a_n} + {b_{n - 1}}$

When $n = 2$ then

${b_2} - {b_1} = {a_2} = 3$

$ \Rightarrow {b_2} - 1 = 3$ [Given ${b_1} = 1$]

$ \Rightarrow {b_2} = 4$

When $n = 3$ then

${b_3} - {b_2} = {a_3}$

$ \Rightarrow {b_3} - 4 = 5$

$ \Rightarrow {b_3} = 9$

$\therefore$ Series is = 1, 4, 9 ......

= 1², 2², 3² ....... n²

$\therefore$ ${b_n} = {n^2}$

Now, $\sum\limits_{n = 1}^{15} {\left( {{a_n}\,.\,{b_n}} \right)} $

$ = \sum\limits_{n = 1}^{15} {\left[ {(2n - 1){n^2}} \right]} $

$ = \sum\limits_{n = 1}^{15} {2{n^3} - \sum\limits_{n = 1}^{15} {{n^2}} } $

$ = 2\left( {{1^3} + {2^3} + \,\,...\,\,{{15}^3}} \right) - \left( {{1^2} + {2^2} + \,\,...\,\,{{15}^2}} \right)$

$ = 2 \times {\left( {{{15 \times 16} \over 2}} \right)^2} - \left( {{{15(16) \times 31} \over 6}} \right)$

$ = 27560$