Question

Solve $\frac{C_0}{1}-\frac{C_1}{5}+\frac{C_2}{9}-\frac{C_3}{13}+...(-1{)}^n\frac{C_n}{4n+1}=\frac{4^{n}n!}{\mathrm{1.5.9....}(4n+1)}$

Answer 1

we required $\frac{C_2}{9}$. Usually we have $C_2{x}^2$ which on integration becomes $C_2\cdots \frac{x^3}{3}$. In order to get $\frac{C_2}{9}$ we must have $C_2(x^4{)}^2=C_2{x}^8$ which on integration becomes $C_2\frac{x^9}{9}.$ Also the terms are alternately + and -. Hence we must have the expansion of $(1-x^4{)}^n$ which is to be integrated within limits 0 to 1.
Now $(1-x^4{)}^n=C_0-C_1{x}^4+C_2{x}^8-C_3{x}^{12}+...+(-1{)}^n{C}_n{x}^{4n}$
Integrate both side within limits 0 to 1 and you get the result.
Let $I_n={\int }_0^1(1-x^4{)}^n$ dx Integrate by parts
$=\left[x\right(1-x^4{)}^{n}dx{]}_0^1-\sum x.n(1-x^4{)}^{n-1}(-4x^3)dx$
$=0-4n\sum (1-x^4{)}^{n-1}(1-x^4-1)dx$
or $I_n=-4n{I}_n+4n{I}_{n-1}or{I}_n(4n+1)=4n{I}_{n-1}$
$\therefore I_n=\frac{4n}{4n+1}{I}_{n-1}=\frac{4n}{4n+1}.\frac{4n-4}{4n-3}\cdots \frac{8}{9}.\frac{4}{5}.I_0$
$=\frac{4^n\left(\mathrm{1.2.3...}n\right)}{\mathrm{1.5.9...}(4n+1)}{I}_0$
where $I_0={\int }_0^1(1-x^4{)}^{0}dx={\int }_0^{1}1dx=1$
Hence from (1) and (2) we get the required result.
$(1+x{)}^n=C_0+C_{1}x+,C_2{x}^2+\cdots +c_r{x}^r+C_n{x}^n$
${(1+\frac{1}{x})}^n=C_0+C_1\frac{1}{x}+C_2\frac{1}{x^2}.+\cdots +C_2\frac{1}{x^r}+\cdots C_n\frac{1}{x^n}$
If we multiply both sides, we get
$\frac{(1+x{)}^{2n}}{x^n}$
$=\sum {C_0}^2+x\sum C_0{C}_1+x^2\sum C_0{C}_2+\cdots x^r\sum C_0{C}_r+\cdots$
The various sums are the coefficients of $x^0,x,x^2,...,x^r$ in L.H.S. $\frac{(1+x{)}^{2n}}{x^n}$
or Coefficients of $x^n,x^{n+1},\cdots ,x^{n+2},\cdots ,x^{n+r}$ in the expansion of $(1+x{)}^{2n}$ which occur in $T_{n+1},T_{n+2},\cdots$ and are ${}^{2n}{C}_{n+1}{,}^{2n}{C}_{n+2,\cdots {,}^{2n}{C}_{n+r}}$ etc.