Question

If $a_0,a_1{,a}_2,...a_n$ is an arithmetic progression of positive integers, then to evaluate

$\frac{C_0}{a_0}\pm \frac{C_1}{a_1}+\frac{C_2}{a_2}\pm ....upto(n+1)terms$, we integrate both the sides of

${\sum }_{r=0}^n{C}_r{x}_r={(1+x)}^n$ after multiplying by $x^r$ for an appropriate value of $r$.

$\frac{1}{3}{C}_0+\frac{1}{4}{C}_1+...upto(n+1)terms$ equal

• A. $\frac{2^{n+1}}{n+3}$
• B. $\frac{2^{n+1}-2^n}{n+4}$
• C. $\frac{2^n}{n+2}$
• D. none of these

Answer 1

The correct option is D none of these

$(1+x{)}^n={}^n{C}_0+{}^n{C}_{1}x+{}^n{C}_2{x}^2+...{}^n{C}_n{x}^n$
Multiplying by $x^2$
$x^2(1+x{)}^n=x^2{}^n{C}_0+{}^n{C}_1{x}^3+{}^n{C}_2{x}^4+...{}^n{C}_n{x}^{n+2}$
Integrating both sides with respect to x, gives us
$\int x^2(1+x{)}^{n}dx=\frac{{}^n{C}_0{x}^3}{3}+\frac{{}^n{C}_1{x}^4}{4}+...$
${\int }_0^1{x}^2(1+x{)}^{n}dx=\frac{{}^n{C}_0}{3}+\frac{{}^n{C}_1}{4}+...$
Let
$I={\int }_0^1{x}^2(1+x{)}^{n}dx$
Let
$1+x=t$
$dt=dx$
Hence
$x^2=(t-1{)}^2$
Therefore
$I={\int }_1^2(t-1{)}^2{t}^{n}dt$
$={\int }_1^2(t^2-2t+1)t^{n}dt$
$={\int }_1^2{t}^{n+2}-2t^{n+1}+t^{n}dt$
$=[\frac{t^{n+3}}{n+3}-\frac{2t^{n+2}}{n+2}+\frac{t^{n+1}}{n+1}{]}_1^2$
$=\frac{2^{n+3}}{n+3}-\frac{{2.2}^{n+2}}{n+2}+\frac{2^{n+1}}{n+1}-\frac{1}{n+3}+\frac{2}{n+2}-\frac{1}{n+1}$
$=\frac{2^{n+1}}{n+1}-\frac{2^{n+3}}{(n+3)(n+2)}+\frac{n}{(n+2)(n+1)}-\frac{1}{n+3}$
Hence
$\frac{2^{n+1}}{n+1}-\frac{2^{n+3}}{(n+3)(n+2)}+\frac{n}{(n+2)(n+1)}-\frac{1}{n+3}=\frac{{}^n{C}_0}{3}+\frac{{}^n{C}_1}{4}+...$