Sequences And Series Question 8
Question 8 - 29 January - Shift 2
Let $a_1=b_1=1$ and $a_n=a _{n-1}+(n-1), b_n=b _{n-1}+a _{n-1}$, $\forall n \geq 2$. If $S=\sum _{n=1}^{10} \frac{b_n}{2^{n}}$ and $T=\sum _{n=1}^{8} \frac{n}{2^{n-1}}$, then $2^{7}(2 S$ $-T$ ) is equal to _________
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Answer: 461
Solution:
Formula: General term ( $n^{\text {th }}$ term) of an A.G.P.
As, $S=\frac{b_1}{2}+\frac{b_2}{2^{2}}+\ldots \ldots .+\frac{b_9}{2^{9}}+\frac{b _{10}}{2^{10}}$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2^{2}}+\frac{b_2}{2^{3}}+\ldots \ldots .+\frac{b_9}{2^{10}}+\frac{b _{10}}{2^{11}}$
Subtracting
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}+(\frac{a_1}{2^{2}}+\frac{a_2}{2^{3}} \ldots \ldots .+\frac{a_9}{2^{10}})-\frac{b _{10}}{2^{11}}$
$\Rightarrow S=b_1-\frac{b _{10}}{2^{10}}+(\frac{a_1}{2}+\frac{a_2}{2^{2}} \ldots \ldots .+\frac{a_9}{2^{9}})$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b _{10}}{2^{11}}+(\frac{a_1}{2^{2}}+\frac{a_2}{2^{3}} \ldots \ldots .+\frac{a_9}{2^{10}})$
Subtracting
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b _{10}}{2^{11}}+(\frac{a_1}{2}-\frac{a_9}{2^{10}})+(\frac{1}{2^{2}}+\frac{2}{2^{3}}+\ldots+\frac{8}{2^{9}})$
$\Rightarrow \frac{S}{2}=\frac{a_1+b_1}{2}-\frac{(b _{10}+2 a_9)}{2^{11}}+\frac{T}{4}$
$\Rightarrow 2 S=2(a_1+b_1)-\frac{(b _{10}+2 a_9)}{2^{9}}+T$
$\Rightarrow 2^{7}(2 S-T)=2^{8}(a_1+b_1)-\frac{(b _{10}+2 a_9)}{4}$
Given $\quad a_n-a _{n-1}=n-1$,
$ \begin{matrix} \therefore \quad & a_2-a_1=1 \\ & a_3-a_2=2 \\ & \vdots \\ \\ & a_9-a_8=8 \\ & a_9-a_1=1+2+\ldots+8=36 \\ a_9=37(a_1=1) \end{matrix} $
Also, $\quad b_n-b _{n-1}=a _{n-1}$
$ \begin{aligned} & \therefore \quad b _{10}-b_1=a_1+a_2+\ldots+a_9 \\ & =1+2+4+7+11+16+22+29+37 \\ & \Rightarrow b _{10}=130(\text{ As } b_1=1. \text{ } \end{aligned} $
