Mathematics - Sequence & Series Question with Solution | TestHub

Given:

$a_n=a_{n-1}+\left(n-1\right)$

$b_n=b_{n-1}+ {\mathrm{a}}_{n-1}$

And,

$S=\frac{b_1}{2}+\frac{b_2}{2^2}+\dots \dots ..+\frac{b_9}{2^9}+\frac{b_{10}}{2^{10}} ...\left(1\right)$

$\Rightarrow \frac{S}{2}= \frac{b_1}{2^2}+\frac{b_2}{2^3}+\dots \dots ..+\frac{b_9}{2^{10}}+\frac{b_{10}}{2^{11}} ...\left(2\right)$

Subtracting $\left(1\right)$ and $\left(2\right)$, we get

$\frac{S}{2}=\frac{b}{2}+\left(\frac{b_2-b_1}{2^2}\right)+\left(\frac{{\displaystyle b_3-b_2}}{{\displaystyle 2^3}}\right)+....+\left(\frac{{\displaystyle b_{10}-b_9}}{{\displaystyle 2^{10}}}\right)-\left(\frac{{\displaystyle b_{10}}}{{\displaystyle 2^{11}}}\right)$

$\Rightarrow \frac{S}{2}=\frac{b_1}{2}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3}\dots \dots .+\frac{a_9}{2^{10}}\right)-\frac{b_{10}}{2^{11}}$

$\Rightarrow S=b_1-\frac{b_{10}}{2^{10}}+\left(\frac{a_1}{2}+\frac{a_2}{2^2}\dots \dots ..+\frac{a_9}{2^9}\right) ....\left(3\right)$

$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b_{10}}{2^{11}}+\left(\frac{a_1}{2^2}+\frac{a_2}{2^3}\dots \dots .+\frac{a_9}{2^{10}}\right) ...\left(4\right)$

$\Rightarrow \frac{S}{2}=\frac{b_1}{2}-\frac{b_{10}}{2^{11}}+\left(\frac{a_1}{2}-\frac{a_9}{2^{10}}\right)+\left(\frac{1}{2^2}+\frac{2}{2^3}+\dots +\frac{8}{2^9}\right)$

$\Rightarrow \frac{S}{2}=\frac{a_1+b_1}{2}-\frac{\left(b_{10}+2a_9\right)}{2^{11}}+\frac{T}{4}$

$\Rightarrow 2S=2\left(a_1+b_1\right)-\frac{\left(b_{10}+2a_9\right)}{2^9}+T$

$\Rightarrow \left(2S-T\right)=2\left(a_1+b_1\right)-\frac{\left(b_{10}+2a_9\right)}{2^9}$

$\Rightarrow 2^7(2S-T)=2^8\left(a_1+b_1\right)-\frac{\left(b_{10}+2a_9\right)}{4}$

Given, 

$a_{\mathrm{n}}-a_{n-1}=n-1$

$a_2-a_1=1$

$a_3-a_2=2$

$$\begin{gathered} ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \end{gathered}$$

$a_9-a_8=8$

Adding all equations, we get

$a_9-a_1=1+2+\dots +8=36$

$\Rightarrow a_9=37 \left(\because a_1=1\right)$

So,

$2^7(2S-T)=2^9-\frac{\left(b_{10}+2·37\right)}{4}$

Also,

$b_n-b_{n-1}=a_{n-1}$

$\therefore b_2-b_1=a_1$

$b_3-b_2=a_2$

$b_4-b_3=a_3$

$$\begin{gathered} ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \\ ⋮ ⋮ ⋮ \end{gathered}$$

$b_{10}-b_9=a_9$

Adding all equations, we get

$b_{10}-b_1=a_1+a_2+...+a_9$

$\Rightarrow b_{10}-b_1=1+2+4+7+11+16+22+29+37$

$\Rightarrow b_{10}=130$

So,

$2^7(2S-T)=2^9-\frac{\left(b_{10}+2·37\right)}{4}=2^9-\frac{\left(130+2·37\right)}{4}=461$