Question

Let $S_n=(2+2)+(2^2+5)+(2^3+10)+(2^4+17)+........$ upto $n$ brackets. If $S_n=2^{n+A}+B{n}^3+C{n}^2+Dn+E$ for all $n\in N$, where $A,B,C,D,E$ are constants then

• A. $B=\frac{1}{6}$
• B. $C=\frac{1}{2}$
• C. $D=1$
• D. $E=-2$

Answer 1

The correct option is D $E=-2$

$S_n=(2+2)+(2^2+5)+(2^3+10)+(2^4+17)+...$

$=(2+2^2+2^3+2^4+...)+(2+5+10+17+...)$

$=\frac{2(2^n-1)}{2-1}+\left[\right(1^2+1)+(2^2+1)+(3^2+1)+...]$ (sum of GP)

$=2^{n+1}-2+\sum _{k=1}^n{K}^2+\sum _{k=1}^{n}K$

$=2^{n+1}-2+\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}$

$=2^{n+1}-2+\frac{n(n+1)(n+2)}{3}$

$S_n=2^{n+1}+\frac{n^3}{3}+n^2+\frac{2}{3}n-2$

On compairing with, $S_n=2^{n+A}+B{n}^3+C{n}^2+Dn+E$

$A=1,B=\frac{1}{3},C=1,D=\frac{2}{3},E=-2$

$\therefore$ Option D is correct