Sequences and Series

$20,19 \frac{1}{4}, 18 \frac{1}{2}, 17 \frac{3}{4}, \ldots \ldots,-129 \frac{1}{4}$

This is A.P. with common difference

$\begin{aligned} & d_1=-1+\frac{1}{4}=-\frac{3}{4} \\ & -129 \frac{1}{4}, \ldots \ldots \ldots \ldots . . .19 \frac{1}{4}, 20 \end{aligned}$

This is also A.P. $\mathrm{a}=-129 \frac{1}{4}$ and $\mathrm{d}=\frac{3}{4}$

Required term $=$

$\begin{aligned} & -129 \frac{1}{4}+(20-1)\left(\frac{3}{4}\right) \\ & =-129-\frac{1}{4}+15-\frac{3}{4}=-115 \end{aligned}$

$4,9,14,19, \ldots$, up to $25^{\text {th }}$ term

$\mathrm{T}_{25}=4+(25-1) 5=4+120=124$

$3,6,9,12, \ldots$, up to $37^{\text {th }}$ term

$\mathrm{T}_{37}=3+(37-1) 3=3+108=111$

Common difference of $\mathrm{I}^{\text {st }}$ series $\mathrm{d}_1=5$

Common difference of $\mathrm{II}^{\text {nd }}$ series $\mathrm{d}_2=3$

First common term $=9$, and their common difference $=15\left(\operatorname{LCM}\right.$ of $\mathrm{d}_1$ and $\left.\mathrm{d}_2\right)$ then common terms are $9,24,39,54,69,84,99$

$ \begin{aligned} & \text t_n=(2 n-1)(2 n+1)(2 n+3) \\\\ & t_n=8 n^3+12 n^2-2 n-3 \\\\ & S_n=\Sigma S_n=8 \Sigma n^3+12 \Sigma n^2-2 \Sigma n-\Sigma 3 \\\\ & =8\left(\frac{n(n+1)}{2}\right)^2+12 \frac{n(n+1)(2 n+1)}{6} -2 \frac{n(n+1)}{2}-3 n \\\\ & =n\left[\frac{8 \cdot n(n+1)^2}{4}+2(n+1)(2 n+1)-n-1-3\right] \\\\ & =n\left[2 n\left(n^2+1+2 n\right)+2\left(2 n^2+3 n+1\right)-n-4\right] \\\\ & =n\left[2 n^3+4 n^2+2 n+4 n^2+6 n+2-n-4\right] \\\\ & =n\left[2 n^3+8 n^2+7 n-2\right] \\\\ & =n(n+1) f(n) \\\\ & \therefore f(n)=\frac{2 n^3+8 n^2+7 n-2}{n+1} \\\\ & \quad f(2)=\frac{16+32+14-2}{3}=\frac{60}{3}=20 \end{aligned} $

The given series is related to the binomial expansion for any index. The binomial expansion for $(1-x)^{-n}$ when $|x| < 1$ is given by :

$ (1-x)^{-n} = 1 + nx + \frac{n(n+1)}{1 \cdot 2}x^2 + \frac{n(n+1)(n+2)}{1 \cdot 2 \cdot 3}x^3 + \ldots $

To relate the series with the binomial expansion, consider setting $x = \frac{1}{2}$ and $n = \frac{2}{3}$. The substitution into the binomial formula results in :

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

This series equates to $\sqrt[3]{4}$ :

$ \sqrt[3]{4} = 1 + \frac{2 \cdot 1}{3 \cdot 2} + \frac{2 \cdot 5 \cdot 1}{3 \cdot 6 \cdot 4} + \ldots $

Thus, the assertion that this infinite series sums to $\sqrt[3]{4}$ is correct, and it is derived from the application of the binomial series expansion with the appropriate substitutions for $x$ and $n$.

We have to check each option, From option (a), we have,

$ \begin{array}{l} (2 n+7) < (n+3)^{2} \\\\ \text { Let } P(n) \equiv(2 n+7) < (n+3)^{2} \\\\ \qquad P(1)=(2+7)=9 < (1+3)^{2}=4^{2}=16 \end{array} $

Let us assume that $ P(K) $ is true.

We have to prove that $ P(K+1) $ is true

$ \begin{array}{l} (2 K+7) < (K+3)^{2} \\\\ 2 K+7 < K^{2}+9+6 K \end{array} $

Add 2 on LHS and $ 7+2 K $ on RHS

as $ 2 < 7+2 K $

$ \begin{array}{l} \Rightarrow 2 K+9 < K^{2}+16+8 K \\\\ 2(K+1)+7 < ((K+1)+3)^{2} \end{array} $

$ \therefore P(K+1) $ is true.

Hence, $ P(x) $ is true, $ \forall n \in N $.

From option b,

$ \begin{array}{r} \text { Let } Q(n) \equiv 1^{2}+2^{2}+\ldots+n^{2} > \frac{n^{3}}{3} \\\\ \qquad Q(1)=1^{2} > \frac{1}{3} \end{array} $

Let $ Q(K) $ be true.

$ \begin{array}{c} 1^{2}+2^{2}+3^{2}+\ldots+K^{2} > \frac{K^{3}}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2} \\\\ +(K+1)^{2} > \frac{K^{3}}{3}+(K+1)^{2} \\\\ 1^{2}+2^{2}+3^{2}+\ldots .+K^{2}+(K+1)^{2} \\\\ > \frac{K^{3}+3 K^{2}+6 K+3}{3} \\\\ 1^{2}+2^{2}+3^{2}+\ldots+K^{2}+(K+1)^{2} > \\\\ \frac{K^{3}+3 K^{2}+3 K+1+3 K+2}{3} \\\\ =\frac{(K+1)^{3}+3 K+2}{3} > \frac{(K+1)^{3}}{3} \end{array} $

$ \therefore Q(K+1)^{2} $ is true.

Hence, $ Q(n) $ is true, $ \forall n \in N $

From option (c),

$ \begin{array}{l} R(n)=35^{2 n+1}+2^{3 n+1} \\\\ R(1)=3 \times 125+2^{4}=375+16=391 \end{array} $

which is divisible by 23

Let us assume it is true for $ n=k, k \in N $

$ \therefore S(k)=3 \cdot 5^{2 k+1}+2^{3 k+1} $ is divisible by

$ 23 \forall k \in N $

Let $ 3 \cdot 5^{2 k+1}+2^{3 k+1}=23 t, t \in N $

$ \therefore 3 \cdot 5 k^{2 k+1}=23 t-2^{3 k+1} $

Now, we have to prove that for $ n=k+1 $

$ \begin{aligned} \therefore S(k+1) & =3 \cdot 5^{2(k+1)+1}+2^{3(k+1)+1} \\\\ & =3 \cdot 5^{2 k+1} \cdot 5^{2}+2^{3 k+1} \cdot 2^{3} \\\\ & =\left(23 t-2^{3 k+1}\right) 25+8 \cdot 2^{3 k+1} \\\\ & =23 t \times 25-25 \times 2^{3 k+1}+8 \cdot 2^{3 k+1} \\\\ \therefore S(k+1) & =23 t \times 25-172^{3 k+1} \end{aligned} $

Hence, $ 3 \cdot 5^{2 n+1}+2^{3 n+1} $ is not divisible by 23.

Hence, $ R(n) $ is not divisible by $ 23, \forall n \in N $. From option (d),

$ 2+7+12+\ldots+(5 n-3) $ is an AP with common difference as 5 .

$ \begin{aligned} S_{n} & =\frac{n}{2}[2 \times 2+(n-1) 5] \\\\ & =\frac{n}{2}(5 n-1) \end{aligned} $

$\begin{aligned} \frac{1}{3 \cdot 6} & +\frac{1}{6 \cdot 9}+\frac{1}{9 \cdot 12}+\ldots \text { to } 9 \text { terms } \\ = & \frac{1}{9}\left(\frac{1}{1 \cdot 2}+\frac{1}{2 \cdot 3}+\frac{1}{3 \cdot 4}+\ldots+\frac{1}{9 \cdot 10}\right) \\ = & \frac{1}{9}\left(\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}-\frac{1}{3}\right)+\left(\frac{1}{3}-\frac{1}{4}\right)\right. \left.+\ldots+\left(\frac{1}{9}-\frac{1}{10}\right)\right) \\ & = \left(1-\frac{1}{10}\right)=\frac{1}{9}\left(\frac{9}{10}\right)=\frac{1}{10}\end{aligned}$

We have,

$ \frac{x}{(x-2)(x-3)}=\frac{A}{(x-2)}+\frac{B}{(x-3)} $

On comparing, we get

$ \begin{aligned} & A=-2 B=3 \\\\ & \Rightarrow \quad \frac{x}{(x-2)(x-3)}=\frac{-2}{x-2}+\frac{3}{x-3} \\\\ &=\frac{1}{\left(1-\frac{x}{2}\right)}+\frac{-1}{\left(1-\frac{x}{3}\right)} \end{aligned} $

$\begin{aligned} & =\left(1-\frac{x}{2}\right)^{-1}-\left(1-\frac{x}{3}\right)^{-1} \\\\ & =\left[1+\frac{x}{2}+\left(\frac{x}{2}\right)^2+\left(\frac{x}{2}\right)^3+\ldots .+\left(\frac{x}{2}\right)^n\right]- \\\\ & \quad\left[1+\frac{x}{3}+\left(\frac{x}{3}\right)^2+\ldots .+\left(\frac{x}{3}\right)^n\right]\end{aligned}$

$\Rightarrow$ Coefficient of $x^2=\frac{1}{4}-\frac{1}{9}=\frac{5}{36}$

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} $

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

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

We know that $ \mathrm{AM} \geq \mathrm{GM} $

Now, applying Eq. (i) in $ a $ and $ b $

$ \Rightarrow \quad \frac{a+b}{2} \geq \sqrt{a b} $

applying Eq. (i) in $ b $ and $ c $

$ \Rightarrow \quad \frac{b+c}{2} \geq \sqrt{b c} $

and applying Eq. (i) in $ c $ and $ a $

$ \Rightarrow \quad \frac{c+a}{2} \geq \sqrt{c a} $

By Eqs. (ii), (iii) and'(iv)

$ \Rightarrow \quad \frac{(a+\bar{b})(b+c)(c+a)}{8} \geq a b c $

$ \Rightarrow \quad(a+b)(b+c)(c+a) \geq 8 a b c $

Let $ S=\sum\limits_{k=1}^{n} k(k+1)(k-1) $

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

We know that

$ \sum\limits_{r=1}^{P} r=\frac{P(P+1)}{2} $

and $ \quad \sum\limits_{r=1}^{P} r^{3}=\left[\frac{P(P+1)}{2}\right]^{2} $

$ \begin{aligned} \therefore & =\left[\frac{n(n+1)}{2}\right]^{2}-\left[\frac{n(n+1)}{2}\right] \\ & =\frac{n(n+1)}{2}\left[\frac{n(n+1)}{2}-1\right] \\ & =\left(\frac{n^{2}+n}{2}\right)\left[\frac{n^{2}+n-2}{2}\right] \\ & =\frac{n^{4}}{4}+\frac{n^{3}}{2}-\frac{n^{2}}{4}-\frac{n}{2} \end{aligned} $

Now, by comparing $ p n^{4}+q n^{3}+t n^{2}+s n $ with Eq. (i),

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

To solve the problem, let's consider the setup described:

The three numbers in an Arithmetic Progression (AP) are denoted as $a$, $a+6$, and $a+12$, where the common difference is 6.

The sequence of four numbers has the first and last numbers equal.

Thus, the four numbers are $(a+12)$, $a$, $a+6$, and $(a+12)$.

The problem states that the first three numbers are in a Geometric Progression (GP). Therefore, the following relationship should hold:

$ a^2 = (a+12)(a+6) $

Expanding and solving the equation:

$ a^2 = a^2 + 18a + 72 $

Simplifying:

$ 0 = 18a + 72 $

Solving for $a$:

$ 18a = -72 $

$ a = -\frac{72}{18} = -4 $

Given this value of $a$, the four numbers are calculated as:

$a+12 = 8$,

$a = -4$,

$a+6 = 2$,

$a+12 = 8$.

Therefore, the sequence of four numbers is $8, -4, 2, 8$.

The two intermediate numbers among the four (excluding the repeated elements at the start and end) are $-4$ and $2$. Hence, the two other numbers in this sequence are $-4, 2$.

Given, $ \frac{e^{7 x}+e^{x}}{e^{3 x}} \Rightarrow e^{4 x}+e^{-2 x} $

Series of $ e^{a}=1+\frac{a}{1!}+\frac{a^{2}}{2!}+\frac{a^{3}}{3!} \ldots $

$ \begin{array}{l} \Rightarrow e^{4 x}+e^{-2 x}=\left(1+\frac{4 x}{1!}+\frac{(4 x)^{2}}{2!}+\frac{(4 x)^{3}}{3!}+\ldots\right) +\left(1+\frac{(-2 x)}{1!}+\frac{(-2 x)^{2}}{2!}+\ldots \ldots\right) \end{array} $

Here, coefficient of

$ \begin{array}{c} x^{2} \equiv \frac{4^{2}}{2!}+\frac{(-2)^{2}}{2!} \\ x^{3} \equiv \frac{4^{3}}{3!}+\frac{(-2)^{3}}{3!} \\ \vdots \\ x^{n}=\frac{4^{n}}{n!}+\frac{(-2)^{n}}{n!} \end{array} $

Now, we have the following relations :

Arithmetic progression :

Since $A_1$ and $A_2$ are arithmetic means between $a$ and $b$, we can say that $a$, $A_1$, $A_2$, and $b$ are in an arithmetic progression. This means there are three equal intervals between $a$ and $b$, which are represented by the common difference $d$.

To find the value of $d$, we can use the following equation :

$ b - a = 3d $

From this equation, we can find the value of $d$ :

$ d = \frac{b - a}{3} $

$ A_1 = a + \frac{b - a}{3} = \frac{2a + b}{3} $

$ A_2 = \frac{a + 2b}{3} $

$ A_1 + A_2 = a + b $

Geometric progression :

$ a, G_1, G_2, G_3, b \text{ are in G.P. } $

$ r = \left(\frac{b}{a}\right)^{\frac{1}{4}} $

$ G_1 = \left(a^3b\right)^{\frac{1}{4}} $

$ G_2 = \left(a^2b^2\right)^{\frac{1}{4}} $

$ G_3 = \left(ab^3\right)^{\frac{1}{4}} $

We have the expression :

$ G_1^4 + G_2^4 + G_3^4 + G_1^2 G_3^2 = a^3b + a^2b^2 + ab^3 + \left(a^3b\right)^{\frac{1}{2}}\cdot\left(ab^3\right)^{\frac{1}{2}} $

Simplify the expression :

$ a^3b + a^2b^2 + ab^3 + ab(a^2b^2) $

Factor out $ab$:

$ ab(a^2 + ab + b^2 + a^2b^2) $

Combine the terms :

$ ab(a^2 + 2ab + b^2) $

Rewrite the expression using the sum of squares :

$ ab(a + b)^2 $

Now, recall that $A_1 + A_2 = a + b$. Substitute this into the expression :

$ G_1 \cdot G_3 \cdot (A_1 + A_2)^2 $

Given the conditions :

1. $a_6 + a_8 = 2 \Rightarrow a r^5 + a r^7 = 2$
2. $a_3 \cdot a_5 = \frac{1}{9} \Rightarrow a^2 \cdot r^2 \cdot r^4 = \frac{1}{9} \Rightarrow a r^3 = \frac{1}{3}$

From this, we can form the equation $\frac{r^2}{3} + \frac{r^4}{3} = 2$, which simplifies to $r^4 + r^2 = 6$.

This can be factored to give $\left(r^2 - 2\right)\left(r^2 + 3\right) = 0$, yielding $r^2 = 2$ (since $r^2$ cannot be $-3$ for real $r$).

So, we have $r = \sqrt{2}$.

Substituting $r = \sqrt{2}$ into the equation $a r = \frac{1}{6}$, we get $a = \frac{1}{6\sqrt{2}}$.

Now, we find the value of $6(a_2+a_4)(a_4+a_6)$:

$6(a_2+a_4)(a_4+a_6) = 6\left(a r + a r^3\right)\left(a r^3 + a r^5\right)$

$= 6\left(\frac{1}{6\sqrt{2}} + \frac{1}{3\sqrt{2}}\right)\left(\frac{1}{3\sqrt{2}} + \frac{2}{3\sqrt{2}}\right)$

$= 6 \cdot \frac{1}{2} \cdot 1 = 3$.