Sequences and Series

We have, $f(n)=A(-2)^n+B(-3)^n$

$ \begin{aligned} f(n)+a f(n-1)+b f(n-2) & =0 \\ \therefore A(-2)^n+B(-3)^n+a\left(A(-2)^{n-1}\right. & \\ \left.+B(-3)^{n-1}\right)+b\left(A(-2)^{n-2}+B(-3)^{n-2}\right. & =0 \\ \Rightarrow A(-2)^{n-2}[4-2 a+b]+(-3)^{n-2} & \\ B[9-3 a+b] & =0 \end{aligned} $

∴ It is possible only

$ 4-2 a+b=0 \text { and } 9-3 a+b=0 $

Solving, we get $a=5, b=6$

$ \therefore \quad(a+b)(b-a)=(5+6)(6-5)=11 $

We have,

$ \begin{array}{r} 1+\frac{\cos \theta}{2}+\frac{\cos 2 \theta}{4}+\frac{\cos 3 \theta}{8}+\ldots . \\ =\frac{a-2 \cos \theta}{5+b \cos \theta} \end{array} $

Put, $\theta=0$

$ \begin{align*} & \therefore & 1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8} \ldots . & =\frac{a-2}{5+b} \\ & \Rightarrow & \frac{1}{1-\frac{1}{2}} & =\frac{a-2}{5+b} \\ & \Rightarrow & 2 & =\frac{a-2}{5+b} \\ & \Rightarrow & 10+2 b & =a-2 \\ & \Rightarrow & a-2 b & =12\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \end{align*} $

Put, $\theta=\pi$

$ 1-\frac{1}{2}+\frac{1}{4}-\frac{1}{8}+\ldots=\frac{a+2}{5-b} $

$ \begin{aligned} &\begin{array}{ll} \Rightarrow & \frac{1}{1+\frac{1}{2}}=\frac{a+2}{5-b} \\ \Rightarrow & \frac{2}{3}=\frac{a+2}{5-b} \\ \Rightarrow & 10-2 b=3 a+6 \\ \Rightarrow & 3 a+2 b=4 \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\end{array}\\ &\text { From Eq. (i) and (ii), we get }\\ &\begin{gathered} a=4, b=-4 \\ \therefore \quad(a-b)^2=(4+4)^2=8^2=64 \end{gathered} \end{aligned} $

We have,

$ \begin{aligned} & \begin{aligned} S_n= & 1^2+2 \times 2^2+3^2+2 \times 4^2+5^2 \\ & +2 \times 6^2+\ldots \infty \end{aligned} \\ \therefore \quad & T_n=(2 n-1)^2+2 \times(2 n)^2 \\ & =4 n^2+1-4 n+8 n^2 \\ = & 12 n^2-4 n+1 \\ & \Sigma T_n=12 \Sigma n^2-4 \Sigma n+\Sigma 1 \\ = & 12 \frac{n(n+1)(2 n+1)}{6}-\frac{4 n(n+1)}{2}+n \\ = & 2 n(n+1)(2 n+1)-2 n(n+1)+n \\ = & 2 n(n+1)(2 n+1-1)+n \\ = & 2 n(n+1)(2 n)+n \\ = & 4 n^2(n+1)+n \\ = & n[4 n(n+1)+1] \\ = & n\left(4 n^2+4 n+1\right)=n(2 n+1)^2 \end{aligned} $

Here, $S_n$ is even

$ \begin{aligned} & S_m=\frac{m}{2}\left(2 \frac{m}{2}+1\right)^2 n=\frac{m}{2} \\ \Rightarrow & S_m=\frac{m}{2}(m+1)^2 \\ \therefore & S_n=\frac{n}{2}(n+1)^2 \text { when } n \text { is even } \end{aligned} $

$ \begin{aligned} &\text { Let roots of the equation }\\ &\begin{aligned} & 8 x^3+6 p x^2+3 q x-27=0 \text { are } \frac{a}{r}, a, \text { ar } . \\ & \therefore \quad \frac{a}{r}+a+a r=\frac{-6 p}{8}=\frac{-3 p}{4} \\ & \Rightarrow \quad a\left(\frac{1}{r}+1+r\right)=\frac{-3 p}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(i) \\ & \left(\frac{a}{r} \cdot a\right)+(a \cdot a r)+\left(\frac{a}{r} \cdot a r\right)=\frac{3 q}{8} \\ & \Rightarrow \quad a^2\left(\frac{1}{r}+r+1\right)=\frac{3 q}{8} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(ii)\\ & \text { and } \quad \frac{a}{r} \times a \times a r=\frac{27}{8} \\ & \Rightarrow \quad a^3=\frac{27}{8} \Rightarrow a=\frac{3}{2} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iii)\end{aligned} \end{aligned} $

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

$ a=\frac{3 q / 8}{-3 p / 4}=-\frac{q}{2 p} \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,...(iv)$

From Eqs. (iii)and (iv), we have

$ \frac{-q}{2 p}=\frac{3}{2} \Rightarrow-q=3 p $

Now,

$ q^2+9 p^2+6 p q+\frac{q}{p} $

$ \begin{array}{ll} \Rightarrow & 9 p^2+9 p^2+6 p(-3 p)+\frac{(-3 p)}{p} \\ \Rightarrow & 18 p^2-18 p^2-3=-3 \end{array} $

$ \begin{aligned} &\\ &\begin{aligned} & \text { We have, } \frac{m}{n}=\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19} \\ & +\ldots+\frac{1}{[1+(20-1) 6][7+(20-1) 6]} \\ & =\frac{1}{1 \cdot 7}+\frac{1}{7 \cdot 13}+\frac{1}{13 \cdot 19}+\ldots+\frac{1}{115 \times 121} \\ & =\frac{1}{6}\left[\frac{6}{1 \cdot 7}+\frac{6}{7 \cdot 13}+\frac{6}{13 \cdot 19}+\ldots+\frac{6}{115 \times 121}\right] \end{aligned} \end{aligned} $

$ \begin{aligned} & =\frac{1}{6}\left[\frac{7-1}{1 \times 7}+\frac{13-7}{7 \times 13}+\frac{19-13}{13 \times 19}+\ldots+\frac{121-115}{115 \times 121}\right] \\ & =\frac{1}{6}\left[\left(\frac{1}{1}-\frac{1}{7}\right)+\left(\frac{1}{7}-\frac{1}{13}\right)+\left(\frac{1}{13}-\frac{1}{19}\right)\right. \\ & \left.+\ldots+\left(\frac{1}{115}-\frac{1}{121}\right)\right] \\ & =\frac{1}{6}\left[1-\frac{1}{121}\right]=\frac{1}{6} \times \frac{120}{121}=\frac{20}{121} \\ & \begin{array}{ll} \therefore & m=20, n=121 \\ \therefore & 5 m+2 n=5 \times 20+2 \times 121 \\ & =100+242=342 \end{array} \end{aligned} $

$\because$ 13, a₁, a₂, ....., a₂₀, 67 are in AP.

$\therefore$ a₁ + a₂ + a₃ + ..... + a₂₀ = 20$\left( {{{13 + 67} \over 2}} \right)$ = 800

Also, AM > GM

$\Rightarrow$ ${{{a_1} + {a_2} + .... + {a_{20}}} \over {20}} \ge {({a_1},{a_2},{a_3},....,{a_{20}})^{{1 \over {20}}}}$

$ \Rightarrow 40 \ge {({a_1}\,.\,{a_2}\,.\,{a_3}\,....\,{a_{20}})^{{1 \over {20}}}}$

Hence, maximum value of ${a_1}\,.\,{a_2}\,.\,{a_3}\,....\,{a_{20}}$ is (40)²⁰.

Since, p, q and r are positive and are in AP.

$\therefore$ $q = {{p + r} \over 2}$ ...... (i)

$\because$ The roots of the equation, px² + qx + r = 0 are real.

$ \Rightarrow {q^2} \ge 4pr \Rightarrow {\left[ {{{p + r} \over 2}} \right]^2} \ge 4pr$ [from Eq. (i)]

$ \Rightarrow {p^2} + {r^2} - 14pr \ge 0$

$ \Rightarrow {\left( {{r \over p}} \right)^2} - 14\left( {{r \over p}} \right) + 1 \ge 0$

$ \Rightarrow {\left( {{r \over p} - 7} \right)^2} - 48 \ge 0$

$ \Rightarrow {\left( {{r \over p} - 7} \right)^2} - {(4\sqrt 3 )^2} \ge 0 \Rightarrow \left| {{r \over p} - 7} \right| \ge 4\sqrt 3 $

Since, $g = \sqrt {ab} $. Also, a, p, q and b are in AP.

So, common difference d is ${{b - a} \over 3}$.

$\therefore$ $p = a + d = a + {{b - a} \over 3} = {{2a + b} \over 3}$

$q = b - d = b - {{b - a} \over 3} = {{a + 2b} \over 3}$

Now, $(2p - q)(p - 2q)$

$ = {{(4a + 2b - a - 2b)} \over 3}\,.\,{{(2a + b - 2a - 4b)} \over 3}$

$ = - ab = - {g^2}$

Since x, y, z are three consecutive integers

$\therefore$ 2y = x + z

$\Rightarrow$ (4y²) = (x + z)²

= (x $-$ z)² + 4xz

= ($-$2)² + 4xz [$\because$ x $-$ z = $-$2]

$\Rightarrow$ 4y² = 4 + 4xz

$\Rightarrow$ y² = 1 + xz

Now,

${1 \over 2}{\log _e}x + {1 \over 2}{\log _e}z + {1 \over {2xz + 1}} + {1 \over 3}{\left( {{1 \over {2xz + 1}}} \right)^3} + ...$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + 2\left( {{1 \over {2zx + 1}} + {1 \over 3}\,.\,{1 \over {{{(2zx + 1)}^3}}} + ...} \right)} \right]$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + {1 \over {2zx + 1}}} \over {1 - {1 \over {2zx + 1}}}}} \right)} \right]$

$ = {1 \over 2}\left[ {{{\log }_e}(xz) + {{\log }_e}\left( {{{1 + xz} \over {xz}}} \right)} \right]$

$ = {1 \over 2}[{\log _e}(1 + xz)] = {1 \over 2}[{\log _e}{(y)^2}] = {\log _e}(y)$

We have, ${S_\infty } = \sum\limits_{k = 1}^\infty {\sum\limits_{n = 1}^\infty {{k \over {{2^{n + k}}}}} } $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\,.\,\sum\limits_{n = 1}^\infty {{1 \over {{2^n}}}} } $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left( {{1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + ....} \right)} $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left[ {{1 \over 2}\left( {1 + {1 \over 2} + {1 \over {{2^2}}}...} \right)} \right]} $

${S_\infty } = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}\left( {{1 \over 2} \times {1 \over {1 - {1 \over 2}}}} \right) = \sum\limits_{k = 1}^\infty {{k \over {{2^k}}}} } $

${S_\infty } = {1 \over 2} + {2 \over {{2^2}}} + {3 \over {{2^3}}} + {4 \over {{2^4}}} + ...$ ..... (i)

${1 \over 2}{S_\infty } = {1 \over {{2^2}}} + {2 \over {{2^3}}} + {3 \over {{2^4}}} + ...$ ...... (ii)

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

${1 \over 2}{S_\infty } = {1 \over 2} + {1 \over {{2^2}}} + {1 \over {{2^3}}} + ...$

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

${1 \over 2}{S_\infty } = 1$

${S_\infty } = 2$