Question

Let $a_1=b_1=1$ and $a_n=a_{n-1}+(n-1),b_n=b_{n-1}+a_{n-1}$, $\forall n\ge 2$. If $S=\sum _{n=1}^{10}\frac{b_n}{2^n}$ and $T=\sum _{n=1}^8\frac{n}{2^{n-1}}$, then $2^7(2S$ $-T)$ is equal to

Answer 1

Answer: 461

As, $S=\frac{b_1}{2}+\frac{b_2}{2^2}+\dots \dots .+\frac{b_9}{2^9}+\frac{b_{10}}{2^{10}}$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2^2}+\frac{b_2}{2^3}+\dots \dots .+\frac{b_9}{2^{10}}+\frac{b_{10}}{2^{11}}$
subtracting
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3}\dots \dots .+\frac{a_9}{2^{10}}\right)-\frac{b_{10}}{2^{11}}$
$\Rightarrow S=b_1-\frac{b_{10}}{2^{10}}+\left(\frac{a_1}{2}+\frac{a_2}{2^2}\dots \dots .+\frac{a_9}{2^9}\right)$
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b_{10}}{2^{11}}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3}\dots \dots .+\frac{a_9}{2^{10}}\right)$
subtracting
$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b_{10}}{2^{11}}+\left(\frac{a_1}{2}-\frac{a_9}{2^{10}}\right)+\left(\frac{1}{2^2}+\frac{2}{2^3}+\dots +\frac{8}{2^9}\right)$
$\Rightarrow \frac{S}{2}=\frac{a_1+b_1}{2}-\frac{\left(b_{10}+2a_9\right)}{2^{11}}+\frac{T}{4}$
$\Rightarrow 2S=2\left(a_1+b_1\right)-\frac{\left(b_{10}+2a_9\right)}{2^9}+T$
$\Rightarrow 2^7(2S-T)=2^8\left(a_1+b_1\right)-\frac{\left(b_{10}+2a_9\right)}{4}$