Question

The value of the series , if $C_0,C_1....C_n$ are Binomial coefficients in $(1+x{)}^n$, then $C_0-\frac{C_1{2}^3}{2}+\frac{C_2{2}^6}{3}-\frac{C_3{2}^9}{4}+...$ up to $(n+1)$ terms equal

• A. $\frac{2^{n+1}-1}{n+1}$
• B. $\frac{1-(-7{)}^{n+1}}{8(n+1)}$
• C. $\frac{1-(-7{)}^{n+1}}{3(n+1)}$
• D. None of these

Answer 1

The correct option is B $\frac{1-(-7{)}^{n+1}}{8(n+1)}$

Let us consider ${(1+x)}^n=C_0+C_{1}x+C_2{x}^2+\cdots +C_n{x}^n\cdots (\ast )$
replacing $x$ by $-x^3$ in $(\ast )$ we get ${(1-x^3)}^n=C_0{x}^0-C_1{x}^3+C_2{x}^6-C_3{x}^9+\cdots$
Multiplying by $x^2$ both sides we get $x^2{(1-x^3)}^n=C_0{x}^2-C_1{x}^5+C_2{x}^8-C_3{x}^{11}+\cdots \left(A\right)$
integrating (A) with respect to $x$ both sides from limits $0$ to $2$, we get
${\int }_0^2{x}^2{(1-3^3)}^{n}dx$ $={\int }_0^2(C_0{x}^2-C_1{x}^5+C_2{x}^8-C_3{x}^{11}+.....)dx$
$x={[\frac{C_0{x}^3}{3}-\frac{C_1{x}^6}{6}+\frac{C_2{x}^9}{9}+\cdots ]}_0^2$ Where $X={\int }_0^2{x}^2{(1-x^3)}^{n}dx$ $=-\frac{1}{3}{\int }_0^2-3x^2{(1-x^3)}^{n}dx$
$=(-\frac{1}{3}){\left[\frac{{(1+x^3)}^{n+1}}{n+1}\right]}_2^0=-\frac{1}{3}\left[\frac{{(-7)}^{n+1}-1}{n+1}\right]$
$=\left[\frac{1-{(-7)}^{n+1}}{3(n+1)}\right]=\frac{C_0{2}^3}{3}-\frac{C_1{2}^6}{6}+\frac{C_2{2}^9}{9}+.....$
$=\left[\frac{1-{(-7)}^{n+1}}{3(n+1)}\right]=\frac{C_0{2}^3}{3}-\frac{C_1{2}^6}{6}+\frac{C_2{2}^9}{9}+.....$
$\frac{1-{(-7)}^{n+1}}{n+1}=3[\frac{C_0{2}^3}{3}-\frac{C_1{2}^6}{6}+\frac{C_2{2}^9}{9}+.....]$
or $=\frac{2^3}{8}(\frac{C_0}{1}-\frac{C_1{2}^3}{2}+\frac{C_2{2}^6}{3}+....+\frac{C_n{2}^{3n}{(-1)}^n}{n})$
by dividing $8$ both sides $\therefore C_0-\frac{C_1{2}^3}{2}+\frac{C_2{2}^6}{3}+....upto(n+1)$ terms $=\frac{-1{(-7)}^{n+1}}{8(n+1)}$