Question

If ${\sum }_{r=0}^{2n}{a}_r{(x-100)}^r={\sum }_{r=0}^{2n}{b}_r{(x-101)}^r$ and $a_k=\frac{2^k}{{{}^{k}C}_n}$ for all $k\ge n$, then $b_n$ equals

• A. $2^n(2^{n+1}-1)$
• B. $2^n(2^n+1)$
• C. $2^n(2^n-1)$
• D. $2^{n+1}(2^n-1)$

Answer 1

The correct option is A $2^n(2^{n+1}-1)$

Substitute $x-101=t$, so that
$\sum _{r=0}^{2n}{b}_r{t}^r=\sum _{r=0}^{2n}{a}_r{(t+1)}^r$ ....(1)
$\therefore b_n=$ coefficient $t^n$ on the $R.H.S$ of (1)
$={{}^{n}C}_n{a}_n+{{}^{n+1}C}_n{a}_{n+1}+.....+{{}^{2n}C}_n{a}_{2n}$
$=\sum _{k=n}^{2n}{{}^{k}C}_r{a}_k=\sum _{k=n}^{2n}{2}^k$
$=2^n\sum _{k=n}^{2n}{2}^{k-n}=2^n(2^{n+1}-1)$