Sequences and Series

As we know,

$ S_n=1^3+2^3+3^3+\ldots+n^3=\left[\frac{n(n+1)}{2}\right]^2 $

and $T_n=1+2+3+\ldots+n$

$ =\frac{n(n+1)}{2} $

Hence, $S_n=T_n^2$

$ \begin{aligned} &\\ &\begin{aligned} & \text { } S_n=\frac{1}{3 \cdot 5}+\frac{1}{5 \cdot 7}+\frac{1}{7 \cdot 9}+\ldots \\ & S_{24}=\sum_{n=1}^{24} \frac{1}{(2 n+1)(2 n+3)} \\ & =\frac{1}{2} \sum_{n=1}^{24} \frac{1}{2 n+1}-\frac{1}{2 n+3} \\ & =\frac{1}{2}\left[\left(\frac{1}{3}-\frac{1}{5}\right)+\left(\frac{1}{5}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{9}\right)\right. \left.+\ldots+\left(\frac{1}{49}-\frac{1}{51}\right)\right] \\ & =\frac{1}{2}\left(\frac{1}{3}-\frac{1}{51}\right)=\frac{1}{2}\left(\frac{16}{51}\right)=\frac{8}{51} . \end{aligned} \end{aligned} $

$1+\frac{4}{15}+\frac{4 \cdot 10}{15 \cdot 30}+\frac{4 \cdot 10 \cdot 16}{15 \cdot 30 \cdot 45}+\ldots+\infty$

We know that

$ \begin{aligned} (1+x)^n=1+n x & +\frac{n(n-1)}{2!} x^2 \\ & +\frac{n(n-1)(n-2)}{3!} x^3+\ldots \end{aligned} $

On comparing with given expansion, we get

$ n x=\frac{4}{15} \Rightarrow x=\frac{4}{15 n} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)$

and $\frac{n(n-1)}{2!} x^2=\frac{4 \cdot 10}{15 \cdot 30}=\frac{40}{450}=\frac{4}{45}$

Using Eq. (i), $\frac{n(n-1)}{2!}\left(\frac{4}{15 n}\right)^2=\frac{4}{45}$

$ \begin{array}{lrl} \Rightarrow & \frac{n(n-1)}{2} \times \frac{4 \times 4}{15 n \times 15 n} & =\frac{4}{45} \\ \Rightarrow & \frac{n(n-1)}{2} \times \frac{16}{225 n^2} & =\frac{4}{45} \\ \Rightarrow & & n=\frac{-2}{3} \end{array} $

$ \begin{aligned} & \text { Also, } x=\frac{4}{15}\left(\frac{-3}{2}\right)=\frac{-2}{5} \\ & \begin{aligned} \therefore \quad(1+x)^n & =\left(1-\frac{2}{5}\right)^{-2 / 3} \\ & =\left(\frac{3}{5}\right)^{-2 / 3}=\left(\frac{5}{3}\right)^{2 / 3} \end{aligned} \end{aligned} $

Given, $t_n=\frac{1}{4}(n+2)(n+3), n \in N \,\,\,\,\,\,\,\,\,\,\,\,\,\ldots$ (i)

Also, $\frac{1}{t_1}+\frac{1}{t_2}+\ldots+\frac{1}{t_{2003}}=\frac{2003}{3009}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)$

From Eq. (i), we get

$ \begin{aligned} \frac{1}{t_n} & =\frac{4}{(n+2)(n+3)} \\ & =\frac{4}{n+2}-\frac{4}{n+3} \end{aligned} $

(By partial fraction)

$ \begin{aligned} \Rightarrow \quad & \sum_{n=1}^{2003} \frac{1}{t_n}=\sum_{n=1}^{2003}\left(\frac{4}{n+2}-\frac{4}{n+3}\right) \\ & =4\left(\frac{1}{3}-\frac{1}{2006}\right)=4\left(\frac{2006-3}{3 \times 2006}\right) \\ & =4 \times \frac{2003}{3 \times 2006} \\ & =\frac{8012}{6018}=\frac{4006}{3009} \end{aligned} $

So, Assertion is false.

Also, by putting $n=2003$ in Reason then we observe that it is wrong.

The integers divisible by 5 from 1 to 100 are $5,10,15,20,25, \ldots, 100$. Which are in A.P. with first term (a) $=5$ and common difference $(d)=5$ ) last term $(l)=100$

So, total number of terms,

$ \begin{aligned} (n) & =\frac{l-a}{d}+1=\frac{100-5}{5}+1=\frac{95}{5}+1 \\ & =19+1=20 \\ \therefore S_n & =\frac{n}{2}(a+l) \\ & =10(5+100)=1050 \end{aligned} $

Similarly, the integers divisible by 13 from 1 to 100 are $13,26,39,52, \ldots, 91$ which are in AP with to $a=13, d=13$, $l=91$.

$ \begin{aligned} & \text { } \begin{array}{l}So,\,\, n=\frac{l-a}{d}+1=\frac{91-13}{13}+1=\frac{78}{13}+1 \\ =6+1=7 \end{array} \\ & \therefore \quad S_n=\frac{7}{2}(13+91)=7 \times 52=364 \end{aligned} $

Now, the integer divisible by both 5 and 13 from 1 to 100 is 65 .

Hence, the sum of integer from 1 to 100 that are divisible by 5 or 13 is $=1050+364-65=1414-65=1349$

Given, $x>\sqrt{3} \Rightarrow x^2>3 \Rightarrow 0<\frac{1}{x^2}<\frac{1}{3}$ Also, $\frac{x^2+1}{\left(x^2+2\right)\left(x^2+3\right)}=\frac{-1}{x^2+2}+\frac{2}{x^2+3} =(-1)\left(x^2+2\right)^{-1}+2\left(x^2+3\right)^{-1}$

$ \begin{aligned} &\begin{aligned} & =(-1)\left(x^2\right)^{-1}\left(1+\frac{2}{x^2}\right)^{-1}+2\left(x^2\right)^{-1}\left(1+\frac{3}{x^2}\right)^{-1} \\ & =\frac{-1}{x^2}\left(1-\frac{2}{x^2}+\frac{4}{x^4}-\frac{8}{x^6}+\frac{16}{x^8}-\ldots \infty\right) \\ & \quad+\frac{2}{x^2}\left(1-\frac{3}{x^2}+\frac{9}{x^4}-\frac{27}{x^6}+\ldots \infty\right) \end{aligned}\\ &\text { So, the coefficient of }\\ &\begin{aligned} x^{-8} & =(-1)(-8)+(2)(-27) \\ & =8-54=-46 \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { From the given, we have }\\ &\begin{aligned} \sum_{k=1}^n k & (k+1)(k+2 \ldots(k+r-1) \\ & =\sum_{K=1}^n \frac{(k+r-1)!}{(k-1)!} \\ & =r!\sum_{k=1}^n{ }^{k+r-1} C_r \\ & =r!{ }^{n+r} C_{r+1} \\ & =\frac{n(n+1)(n+2) \ldots(n+r)}{(r+1)} \end{aligned} \end{aligned} $

$ \begin{aligned} &\text { } \mathrm{AM} \geq \mathrm{GM}\\ &\begin{array}{lc} \therefore & \frac{1+3+3^2+3^3+3^{n-1}}{n} \\ & \geq\left(1 \cdot 3 \cdot 3^2 \cdot 3^3 \cdot \ldots \cdot 3^{n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2 n} \geq\left(3^{1+2+3+\ldots n-1}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n\left(3^n \frac{(n-1)}{2}\right)^{\frac{1}{n}} \\ \Rightarrow & \frac{3^n-1}{2} \geq n 3^{\frac{n-1}{2}} \end{array} \end{aligned} $

$ \begin{aligned} &\text { Given, } 2 \cdot 5+5 \cdot 9+8 \cdot 13+11 \cdot 17+\ldots+\text { upto } n \text { terms }\\ &\begin{aligned} & t_n=(3 n-1)(4 n+1) \\ & t_n=12 n^2-n-1 \\ & \therefore S_n=\Sigma t_n=\Sigma\left(12 n^2-n-1\right)=12 \Sigma n^2-\Sigma n-\Sigma 1 \\ & =12 \cdot \frac{n(n+1)(2 n+1)}{6}-\frac{n(n+1)}{2}-n \\ & =n\left[2\left(2 n^2+3 n+1\right)-\frac{n+1}{2}-1\right]=\frac{n}{2}\left[8 n^2+12 n+4-n-1-2\right] \\ & =\frac{n}{2}\left[8 n^2+11 n+1\right] \\ & =4 n^3+\frac{11}{2} n^2+\frac{n}{2}=a n^3+b n^2+c n+d \\ & \Rightarrow a=4, b=\frac{11}{2}, c=\frac{1}{2} \text { and } d=0 \\ & \therefore a-b+c-d=4-\frac{11}{2}+\frac{1}{2}-0=4-5=-1 \end{aligned} \end{aligned} $

$ \begin{aligned} & \text { We have, } 1^3+2^3+3^3+\ldots+n^3>x \\ & \because \quad \Sigma n^3=\left[\frac{n(n+1)}{2}\right]^2=\frac{n^2}{4}(n+1)^2>x \\ & \because \quad \frac{n^2}{4}<\frac{n^2}{4}(n+1)^2 \\ & \therefore \quad x=\frac{n^2}{4} \end{aligned} $

We have,

$ 1+(3+5+7)+(9+11+13+15+17)+\ldots $

Total number of terms

$ \begin{aligned} & =1+3+5+7+\ldots+n \text { terms } \\ & =\frac{n}{2}[2+(n-1) 2] \\ & =n^2 \\ S_n=1 & +3+5+7+9+11+13+15+17+\ldots \end{aligned} $

upto $n^2$ terms

$ \begin{aligned} S_n & =\frac{n^2}{2}\left[2+\left(n^2-1\right) 2\right] \\ & =\frac{n^2}{2} \times\left(2 n^2\right)=n^4 \\ a_n & =S_n-S_{n-1} \\ & =n^4-(n-1)^4 \\ & =(n-(n-1))(n+(n-1))\left(n^2+(n-1)^2\right) \\ & =(2 n-1)\left[n^2+(n-1)^2\right] \end{aligned} $

Here, a i.e. first term $\geq 1$

$a r^2($ third term $) \leq 100$

$ \begin{aligned} & r^2 \leq 100 \\ \Rightarrow \quad & r \leq 10 \end{aligned} $

When $r=2$, then $a \leq \frac{100}{4}=25$

$\therefore a$ can have 25 values

When $r=3$, then $a \leq \frac{100}{9} \Rightarrow a \leq 11$

$\therefore a$ can have 11 values

When $r=4$, then $a \leq \frac{100}{16} \Rightarrow a \leq 6$

$\therefore a$ can have 6 values.

When $r=5$, then $a \leq \frac{100}{25}=4$

$\therefore a$ can have 4 values.

When $r=6$, then $a \leq \frac{100}{36} \Rightarrow a \leq 2$

$\therefore a$ can have 2 values.

When $r=7$, then $a \leq \frac{100}{49} \Rightarrow a \leq 2$

$\therefore a$ can have 2 values.

When $r=8,9$ and 10 , then

$a \leq \frac{100}{64}, a \leq \frac{100}{81}$ and $a \leq \frac{100}{100}$,

respectively.

$\therefore a$ can have 1 value in each case. Hence, total number of ways.

$ \begin{aligned} & =25+11+6+4+2+2+1+1+1 \\ & =53 \end{aligned} $

$2+3+5+6+8+9 \ldots 2 n$ terms

Observe the sequence and note that every third term is missing.

So, we can split this summation into two Arithmatic Progiression (AP's).

First $\quad A P_1=2+5+8+\ldots n$ terms

Then, $\quad S_1=\frac{n}{2}\left[2 a_1+(n-1) d_1\right]$

Similarly, $A P_2=3+6+9+\ldots n$ terms

Then, $\quad S_2=\frac{n}{2}\left[2 a_2+(n-1) d_2\right]$

Therefore, $S=S_1+S_2$

$ \begin{aligned} S= & \frac{n}{2}[2 \times 2+(n-1) 3] +\frac{n}{2}[2 \times 3+(n-1) \times 3] \\ = & \frac{n}{2}[4+3(n-1)+6+3(n-1)] \\ = & \frac{n}{2}[10+3 n-3+3 n-3] \\ = & \frac{n}{2}[4+6 n]=\frac{n}{2}[2+3 n] \times 2 \\ = & 2 n+3 n^2 \\ S & =3 n^2+2 n \end{aligned} $

When solving the quadratic equation $x^2 - 6x - 2 = 0$, the roots $\alpha$ and $\beta$ can be found using the quadratic formula:

$ \alpha, \beta = \frac{6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot (-2)}}{2} = \frac{6 \pm \sqrt{36 + 8}}{2} = \frac{6 \pm \sqrt{44}}{2} = 3 \pm \sqrt{11} $

Given that $a_n = \alpha^n - \beta^n$ for $n \geq 1$, and knowing $\alpha$ and $\beta$ are roots, they satisfy:

$ t^2 - 6t - 2 = 0 \implies t^2 = 6t + 2 $

Multiplying by $t^{n-2}$, we have:

$ t^n = 6t^{n-1} + 2t^{n-2} $

Therefore, for $\alpha$:

$ \alpha^n = 6\alpha^{n-1} + 2\alpha^{n-2} $

And similarly for $\beta$:

$ \beta^n = 6\beta^{n-1} + 2\beta^{n-2} $

This gives us for $a_n$:

$ \begin{aligned} a_n &= \alpha^n - \beta^n \\ &= 6(\alpha^{n-1} - \beta^{n-1}) + 2(\alpha^{n-2} - \beta^{n-2}) \\ a_n &= 6a_{n-1} + 2a_{n-2} \end{aligned} $

To find $\frac{a_{10} - 2a_8}{2a_9}$, we apply the recurrence relation:

Given $a_{10} = 6a_9 + 2a_8$, we have:

$ a_{10} - 2a_8 = 6a_9 + 2a_8 - 2a_8 = 6a_9 $

Thus,

$ \frac{a_{10} - 2a_8}{2a_9} = \frac{6a_9}{2a_9} = 3 $

Therefore, the value is $3$.

$ \begin{aligned} & \frac{x^4}{(x-2)(x+1)}=x^2+x+3+\frac{-\frac{1}{3}}{x+1}+\frac{\frac{16}{3}}{x-2} \\ & =x^2+x+3+\left(-\frac{1}{3}\right)(1+x)^{-1}+\frac{16}{3}(x-2)^{-1} \end{aligned} $

$ \begin{array}{r} =x^2+x+3-\frac{1}{3}\left[1-x+x^2-x^3+\ldots\right] \\ -\frac{8}{3}\left(1-\frac{x}{2}\right)^1 \end{array} $

$ \begin{aligned} &\begin{aligned} =x^2+x+3- & \frac{1}{3}\left[1-x+x^2-x^3+\ldots\right] \\ & -\frac{8}{3}\left[1+\frac{x}{2}+\frac{x^2}{4}+\ldots\right] \end{aligned}\\ &\text { Hence, coefficient of } x^2 \text { is }\left(1-\frac{1}{3}-\frac{2}{3}\right)\\ &=(1-1)=0 \end{aligned} $

To solve the given problem, we need to analyze the expression for the sum of terms and match it with the provided equation.

The sequence is structured as:

$ 1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots $

We define the $ r $-th term as:

$ a_r = (2r-1)(2r+1)(2r+3) $

Calculating $ a_r $, we expand:

$ a_r = (2r-1)(2r+1)(2r+3) = 8r^3 + 12r^2 - 2r - 3 $

The sum $ S_n $ of the first $ n $ terms is:

$ S_n = \sum_{r=1}^{n} a_r = \sum_{r=1}^{n} \left( 8r^3 + 12r^2 - 2r - 3 \right) $

Breaking it down, we compute:

$ S_n = 8 \sum_{r=1}^{n} r^3 + 12 \sum_{r=1}^{n} r^2 - 2 \sum_{r=1}^{n} r - 3 \sum_{r=1}^{n} 1 $

Substituting the formula for each sum, we have:

$ 8 \left[ \frac{n(n+1)}{2} \right]^2 + 12 \left[ \frac{n(n+1)(2n+1)}{6} \right] - 2 \left[ \frac{n(n+1)}{2} \right] - 3n $

Simplifying, this becomes:

$ S_n = n(n+1)(2n^2 + 6n + 1) - 3n $

Here, the function $ f(n) $ is:

$ f(n) = 2n^2 + 6n + 1 $

To find $ f(1) $, substitute $ n = 1 $:

$ f(1) = 2(1)^2 + 6(1) + 1 = 9 $

Thus, the function $ f(l) $ evaluates to $ 9 $.

Given the polynomial equation $ x^3 - bx^2 + cx - d = 0 $, we need to find the condition for its roots to be in arithmetic progression.

Let the roots of this polynomial be $\alpha$, $\beta$, and $\gamma$. Since these roots are in arithmetic progression, they satisfy the condition:

$ 2\beta = \alpha + \gamma $

We also know from the properties of polynomials that the sum of the roots is equal to the coefficient of the $x^2$ term, which is $b$. Thus,

$ \alpha + \beta + \gamma = b $

Substituting the condition of arithmetic progression:

$ 3\beta = b \quad \Rightarrow \quad \beta = \frac{b}{3} $

Next, we use the condition for the sum of the products of the roots taken two at a time:

$ \alpha \beta + \beta \gamma + \gamma \alpha = c $

And for the product of the roots:

$ \alpha \beta \gamma = d $

Substituting $\beta = \frac{b}{3}$, the product becomes:

$ \alpha \gamma \cdot \frac{b}{3} = d \quad \Rightarrow \quad \alpha \gamma = \frac{3d}{b} $

Now substituting the values of $\beta$ and $\alpha \gamma$ into the equation for the sum of the products of the roots taken two at a time:

$ \alpha \cdot \frac{b}{3} + \frac{b}{3} \gamma + \frac{3d}{b} = c $

Simplifying further:

$ \frac{b}{3} (\alpha + \gamma) + \frac{3d}{b} = c $

Using the condition $ \alpha + \gamma = 2\beta $:

$ \frac{b}{3} \cdot 2\beta + \frac{3d}{b} = c $

Substituting $\beta = \frac{b}{3}$:

$ \frac{b}{3} \cdot \frac{2b}{3} + \frac{3d}{b} = c $

Simplifying this expression leads to:

$ \frac{2b^2}{9} + \frac{3d}{b} = c $

By multiplying through by $9b$ to clear denominators, we become:

$ 2b^3 + 27d = 9bc $

Thus, the condition for the roots of the polynomial to be in arithmetic progression is:

$ 9bc = 2b^3 + 27d $

Given expression $\frac{2 x+1}{(1+x)(1-2 x)}$

By partial fraction

$ \begin{aligned} \frac{(2 x+1)}{(1+x)(1-2 x)} & =\frac{A}{(1+x)}+\frac{B}{(1-2 x)} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i)\\ 2 x+1 & =A(1-2 x)+B(1+x) \\ 2 x+1 & =A+B+(-2 A+B) x \end{aligned} $

On comparing coefficients both sides, we get

$ \begin{gathered} A+B=1 \Rightarrow A=1-B \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, -2 A+B=2 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{gathered} $

From Eqs. (ii) and (iii), we get

$ \begin{aligned} & & -2+2 B+B & =2 \\ \Rightarrow & & B & =\frac{4}{3} \text { and } A=-\frac{1}{3} \end{aligned} $

On putting the value of $A$ and $B$ in Eq.

(i), we get

$ \begin{aligned} & \frac{2 x+1}{(1+x)(1-2 x)}=-\frac{1}{3(1+x)}+\frac{4}{3(1-2 x)} \\ & =-\frac{1}{3}(1+x)^{-1}+\frac{4}{3}(1-2 x)^{-1} \end{aligned} $

$ \begin{aligned} & =-\frac{1}{3}\left[1-x+x^2-x^3+x^4\right.\left.\quad-x^5+x^6-x^7+x^8-x^9\right ] +\frac{4}{3}\left[1+2 x+4 x^2+8 x^3+16 x^4+32 x^5\right] \\ & \left.64 x^6+128 x^7+256 x^8+512 x^9+\ldots\right] \end{aligned} $

$ \begin{aligned} & =\left(-\frac{1}{3}+\frac{4}{3}\right)+\left(\frac{1}{3}+\frac{1}{8}\right) x+\left(-\frac{1}{3}+\frac{16}{3}\right) x^2 +\left(\frac{1}{3}+\frac{32}{3}\right) x^3+\left(-\frac{1}{3}+\frac{64}{3}\right) x^4+\left(\frac{1}{3}+\frac{128}{3}\right) x^5 \\ & \quad+\left(-\frac{1}{3}+\frac{256}{3}\right) x^6+\left(\frac{1}{3}+\frac{512}{3}\right) x^7 \quad+\left(-\frac{1}{3}+\frac{1024}{3}\right) x^8+\left(\frac{1}{3}+\frac{2024}{3}\right) x^5 \end{aligned} $

Now, sum of coefficients of first 5 odd power of $x$, we get

$ \begin{aligned} & =\left(\frac{1}{3}+\frac{8}{3}\right)+\left(\frac{1}{3}+\frac{32}{3}\right)+\left(\frac{1}{3}+\frac{128}{3}\right)+\left(\frac{1}{3}+\frac{512}{3}\right)+\left(\frac{1}{3}+\frac{2048}{3}\right) \\ & =\left[\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}\right]\quad+\left[\frac{8}{3}+\frac{32}{3}+\frac{128}{3}+\frac{512}{3}+\frac{2048}{3}\right] \end{aligned} $

$ \begin{aligned} & =\frac{5}{3}+\frac{1}{3}[2728] \\ & =\frac{5}{3}+\frac{8}{9}(1023) \\ & =\frac{5}{3}+\frac{8}{9}\left(4^5-1\right) \end{aligned} $

To understand the summation of the given series, consider the series:

$ \frac{1}{1 \cdot 5} + \frac{1}{5 \cdot 9} + \frac{1}{9 \cdot 13} + \ldots \text{ up to } n \text{ terms} $

We denote this sum as $ S $:

$ S = \frac{1}{1 \cdot 5} + \frac{1}{5 \cdot 9} + \frac{1}{9 \cdot 13} + \ldots + \frac{1}{(4n-3)(4n+1)} $

We can express each term in a telescoping form:

$ S = \frac{1}{4} \left\{ \frac{5-1}{1 \cdot 5} + \frac{9-5}{5 \cdot 9} + \frac{13-9}{9 \cdot 13} + \ldots + \frac{(4n+1) - (4n-3)}{(4n-3)(4n+1)} \right\} $

Simplifying further using partial fractions, we have:

$ = \frac{1}{4} \left[ \left( 1 - \frac{1}{5} \right) + \left( \frac{1}{5} - \frac{1}{9} \right) + \left( \frac{1}{9} - \frac{1}{13} \right) + \ldots + \left( \frac{1}{4n-3} - \frac{1}{4n+1} \right) \right] $

Observe that it forms a telescoping series, where most middle terms cancel out. Thus, we are left with:

$ = \frac{1}{4} \left[ 1 - \frac{1}{(4n+1)} \right] $

Finally, we express this as:

$ = \frac{1}{4} \left[ \frac{4n+1 - 1}{4n+1} \right] = \frac{1}{4} \cdot \frac{4n}{4n+1} = \frac{n}{4n+1} $

Thus, the sum of the series up to $ n $ terms is:

$ \frac{n}{4n+1} $

To solve for $ m $ in the equation $ 4x^3 - 12x^2 + 11x + m = 0 $, given that the roots are in arithmetic progression, we proceed as follows:

Assume the roots of the polynomial are $ a-d $, $ a $, and $ a+d $.

Sum of Roots:

According to the polynomial equation, the sum of the roots is given by:

$ (a-d) + a + (a+d) = \frac{-b}{a} = \frac{12}{4} = 3 $

Simplifying the left-hand side:

$ 3a = 3 $

Solving for $ a $, we find:

$ a = 1 $

Using the Root in the Original Equation:

Since $ a = 1 $ is a root of the polynomial, it should satisfy the equation:

$ 4(1)^3 - 12(1)^2 + 11(1) + m = 0 $

Simplify this:

$ 4 - 12 + 11 + m = 0 $

$ 3 + m = 0 $

Solving for $ m $:

$ m = -3 $

Thus, the value of $ m $ is $-3$.

Given, $P(n)=2 \cdot 4^{2 n+1}+3^{3 n+1}$

For $n=1$

$ \begin{aligned} P(1) & =2 \cdot 4^{2+1}+3^{3+1}=2 \cdot 4^3+3^4 \\ & =2 \cdot 64+81=128+81=209 \end{aligned} $

For $n=2$

$ \begin{aligned} P(2) & =2 \cdot 4^{4+1}+3^{6+1}=2 \cdot 4^5+3^7 \\ & =2048+2187=4235 \end{aligned} $

Now, HCF of 209 and 4235 is 11.

$\because 2 \cdot 4^{2 n+1}+3^{3 n+1}$ is divisible by 11 ,

$\forall n \in N$.

$ \therefore \quad k=11 $

To solve the problem where the roots of the polynomial equation $ x^3 + ax^2 + bx + c = 0 $ are in arithmetic progression, let's denote these roots as $ s-t, s, $ and $ s+t $.

Sum of Roots: From Vieta's formulas, we know the sum of the roots is equal to $-a$. Therefore,

$ (s-t) + s + (s+t) = -a $

Simplifying gives:

$ 3s = -a \quad \Rightarrow \quad a = -3s $

Sum of Products of Roots Taken Two at a Time: The sum of the products of the roots taken two at a time is equal to $b$. Thus,

$ (s-t)s + s(s+t) + (s-t)(s+t) = b $

This expands to:

$ s^2 - st + s^2 + st + s^2 - t^2 = b $

Simplifying gives:

$ 3s^2 - t^2 = b $

Product of Roots: The product of the roots is equal to $-c$. Therefore,

$ (s-t)s(s+t) = -c $

This simplifies to:

$ s^3 - st^2 = -c $

So, we have:

$ c = s^3 - st^2 $

Calculating $2a^3 + 27c$ and $9ab$:

$ 2a^3 + 27c = 2(-3s)^3 + 27(-s^3 + st^2) $

Calculating gives:

$ = 2(-27s^3) + 27(-s^3 + st^2) $

$ = -54s^3 - 27s^3 + 27st^2 $

$ = -81s^3 + 27st^2 $

For $9ab$:

$ 9ab = 9(-3s)(3s^2 - t^2) $

$ = -81s^3 + 27st^2 $

Therefore,

$ 9ab = 2a^3 + 27c $

The equation $ 9ab = 2a^3 + 27c $ holds true when the roots of the polynomial are in arithmetic progression.

$\frac{1}{3 \cdot 7}+\frac{1}{7 \cdot 11}+\frac{1}{11 \cdot 15}+\ldots$ to 50 terms

We can see that each term in the series can be represented as $\frac{1}{(4 n-1)(4 n+3)}$

So, $S=\sum\limits_{n=1}^{50} \frac{1}{(4 n-1)(4 n+3)}$

Using partial fraction

$ S=\sum\limits_{n=1}^{50} \frac{1}{4}\left[\frac{1}{4 n-1}-\frac{1}{4 n+3}\right] $

$ \begin{aligned} & =\frac{1}{4}\left[\frac{1}{3}-\frac{1}{7}+\frac{1}{7}-\frac{1}{11}+\frac{1}{11}-\frac{1}{15} \cdots \frac{1}{199}-\frac{1}{203}\right] \\ & =\frac{1}{4}\left[\frac{1}{3}-\frac{1}{203}\right]=\frac{1}{4}\left[\frac{203-3}{609}\right]=\frac{1}{4}\left[\frac{200}{609}\right]=\frac{50}{609} \end{aligned} $

We are given the infinite series:

$ 1 + \frac{1}{3} + \frac{1 \cdot 3}{3 \cdot 6} + \frac{1 \cdot 3 \cdot 5}{3 \cdot 6 \cdot 9} + \ldots $

This series can be related to the binomial expansion of $(1 + x)^n$, which is given by:

$ (1+x)^n = 1 + nx + \frac{n(n-1)x^2}{2!} + \ldots $

Comparing this with the terms of our series, we identify:

$ nx = \frac{1}{3} $ implies $ x = \frac{1}{3n} $.

For the second comparison, $\frac{n(n-1)x^2}{2} = \frac{1}{6}$.

Substitute $ x = \frac{1}{3n} $ into the second equation:

$ \frac{n(n-1)}{2} \times \left(\frac{1}{3n}\right)^2 = \frac{1}{6} $

Simplifying gives:

$ \frac{n(n-1)}{18n^2} = \frac{1}{6} $

$ \frac{n-1}{3n} = 1 $

This simplifies to:

$ 3n - n = -1 $

$ 2n = -1 $

$ n = -\frac{1}{2} $

Substituting back to find $ x $:

$ x = \frac{1}{3} \times \frac{-2}{1} = \frac{-2}{3} $

Thus, the series becomes:

$ (1+x)^n = \left(1 - \frac{2}{3}\right)^{-1/2} = \left(\frac{1}{3}\right)^{-1/2} = 3^{1/2} = \sqrt{3} $

Therefore, the sum of the series is $\sqrt{3}$.

We have,

$ 2 \cdot 5+5 \cdot 9+8 \cdot 13+11 \cdot 17+\ldots+\text { upto } 10 $

terms.

$ \begin{aligned} & =(3-1)(4+1)+(6-1)(8+1)+(9-1) \\ & (12+1)+(12-1)(16+1)+\ldots+\text { upto } 10 \end{aligned} $

terms.

$ \begin{aligned} & =\sum_{r=1}^{10}(3 r-1)(4 r+1) \\ & =\sum_{r=1}^{10}\left(12 r^2-r-1\right) \\ & =12 \times \frac{10 \times 11 \times 21}{6}-\frac{10 \times 11}{2}-10 \\ & =4620-55-10=4555 \end{aligned} $

We have, the roots of $x^3-13 x^2+k x$ $-27=0$ are in GP be.

Let the roots of equation be $\frac{a}{r}, a$, $a r$

$ \begin{aligned} & \frac{a}{r} \times a \times a r=27 \Rightarrow a^3=27 \Rightarrow a=3 \\ & \Rightarrow \quad \frac{a}{r}+a+a r=13 \\ & \Rightarrow \quad \frac{3}{r}+3+3 r=13 \\ & \Rightarrow \quad 3\left(r+\frac{1}{r}\right)=10 \\ & \Rightarrow \quad 3 r^2-10 r+3=0 \\ & \Rightarrow \quad 3 r^2-9 r-r+3=0 \\ & \Rightarrow \quad(r-3)(3 r-1)=0 \\ & \Rightarrow \quad r=3, \frac{1}{3} \end{aligned} $

$\therefore$ The roots of the equation are 1,3 and 9 .

$ \begin{aligned} & K=1 \times 3+3 \times 9+9 \times 1=3+27+9 \\ & =39 \end{aligned} $

$\begin{aligned} & \ \begin{array}{l}\text { Let } S_{\infty}=1-\frac{2}{3}+\frac{2 \cdot 4}{3 \cdot 6}-\frac{2 \cdot 4 \cdot 6}{3 \cdot 6 \cdot 9}+\ldots \infty \\ =\left(1-\frac{2}{3}\right)+\frac{2 \cdot 4}{3 \cdot 6}\left(1-\frac{6}{9}\right)+\frac{2 \cdot 4 \cdot 6 \cdot 8}{3 \cdot 6 \cdot 9 \cdot 12} \\ \left(1-\frac{10}{15}\right)+\ldots \infty\end{array}\end{aligned}$

$\begin{aligned} & =\left(1-\frac{2}{3}\right)\left[1+\frac{2 \cdot 4}{3 \cdot 6}+\frac{2 \cdot 4 \cdot 6 \cdot 8}{3 \cdot 6 \cdot 9 \cdot 12}+\ldots \infty\right] \\ & =\frac{1}{3}\left[1+\left(\frac{2}{3}\right)^2+\left(\frac{2}{3}\right)^4+\ldots \infty\right] \\ & =\frac{1}{3}\left[\frac{1}{1-\left(\frac{2}{3}\right)^2}\right]=\frac{1}{3} \times \frac{9}{5}=\frac{3}{5}\end{aligned}$

Let the coordinate $A\left(x_1, y_1\right), B\left(x_2, y_2\right)$ and $C\left(x_3, y_3\right)$.

According to question, $x$ - coordinate as well as $y$ coordinates are in GP.

Let $x_1=a, x_2=a r$ and $x_3=a r^2,y_1=b, y_2=b r$ and $y_3=b r^2$

Now, $A(a, b), B(a r, b r), C\left(a r^2, b r^2\right)$.

Slope of $A B=\frac{b(1-r)}{a(1-r)}=\frac{b}{a}$ and

Slope of $\mathrm{BC}=\frac{b r(1-r)}{\operatorname{ar}(1-r)}=\frac{b}{a}$

As slope of $A B=$ Slope of $B C$

$\therefore$ A, B and C are collinear.

$\therefore$ A, B and C lie on a straight line.

Given, $a_0=1, a_{n+1}=3 n^2+n+a_n$ .... (i)

From Eq.(i), we get

$\begin{array}{ll} & a_1=3(0)^2+0+a_0=1 \\ \Rightarrow \quad & a_1=1 \\ & a_2=3(1)^2+1+a_1=3+1+1=5 \\ \Rightarrow \quad & a_2=5 \end{array}$

$\begin{aligned} & a_3=3(2)^2+2+a_2=12+2+5=19 \\ & \Rightarrow \quad a_3=19 \\ \end{aligned}$

Now, check the option for $n=3$ which expression satisfy by $a_3=19$

Option (a), $a_n=n^3+n^2+1$

Put $n=3$, then

$\begin{array}{rlrl} a_3 & =(3)^3+(3)^2+1 \\ & =27+9+1=37 \\ \because \quad a_3 & \neq 37 \\ \therefore \quad a_n & \neq n^3+n^2+1 \\ & \text { Option (b), } a_n =n^3-n^2+1 \end{array}$

Option (b), $a_n=n^3-n^2+1$

Put $n=3$, then

$\begin{aligned} a_3 & =(3)^3-(3)^2+1 \\ & =27-9+1=19 \\ \therefore \quad &a_n =n^3-n^2+1 \end{aligned}$

$\begin{aligned} & \text { Given, } x^2-\sqrt{3} x+1=0 \\ & \Rightarrow \quad x=\frac{\sqrt{3} \pm \sqrt{3-4}}{2} \\ & x=\frac{\sqrt{3}}{2} \pm \frac{1}{2} i=\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{2 n \pi}{6} \pm i \sin \frac{2 n \pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \text { and } x^{-2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \operatorname{Now},\left(x^n-\frac{1}{x^n}\right)^2=\left(x^{2 n}+\frac{1}{x^{2 n}}-2\right) \\ & =-2+2 \cos \frac{n \pi}{3} \\ & \Rightarrow \sum_\limits{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)^2=\sum_\limits{n=1}^{24}\left(-2+2 \cos \frac{n \pi}{3}\right) \\ & =-48+2\left[\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\ldots+\cos \frac{24 \pi}{3}\right] \\ & =-48+2\left[\frac{\cos \left(\frac{2 \pi}{3}+\frac{23 \pi}{6}\right) \cdot \sin \frac{24 \pi}{6}}{\sin \frac{\pi}{6}}\right] \\ & =-48+2 \cdot 0=-48 \\ & \end{aligned}$

$\because$ $p,q,r,s$ are in AP.

Let $p = a,q = a + d,r = a + 2d,s = a + 3d$

Given, $p,q$ are roots of

${x^2} - 2x + A = 0$

Then, $p + q = 2$ and $pq = A$

$ \Rightarrow 2a + d = 2$ .... (i)

Also, r and s are roots of ${x^2} - 18x + B = 0$

Then, $r + s = 18$ and $rs = B$

$ \Rightarrow 2a + 5d = 18$ ..... (ii)

Solving Eqs. (i) and (ii), we get

$a = - 1,d = 4$

Now, $pq = A$ (product of roots)

or $A = a(a + d) = ( - 1)( - 1 + 4)$

$A = - 3$

and $rs = B$

or $B = (a + 2d)(a + 3d)$

$ = ( - 1 + 8)( - 1 + 12) = 77$

$\therefore$ $A = - 3,B = 77$

$f(x)=x^3+a x^2+b x+c$

Let roots of $f(x)$ are $\alpha, \beta, \gamma$ and since roots are in AP.

$\therefore \quad 2 \beta=\alpha+\gamma$

Now, Sum of roots

$\begin{aligned} \alpha+\beta+\gamma & =-a \\ \text { or }2 \beta+\beta & =-a \text { or } \beta=-\frac{a}{3} \end{aligned}$

It is given that, roots are integers.

$\therefore \beta$ is an integer when $a$ is multiple of 3 .

Option (a), (c) and (d) are multiple of 3.

$\Rightarrow \quad a \neq 1214$

Option (b) is correct.