Question

Prove independently that
$\frac{C_0}{1}-\frac{C_1}{4}+\frac{C_2}{7}-.......+(-1{)}^n\frac{C_n}{3n+1}=\frac{3^n\cdot n!}{\mathrm{1.4.7.....}(3n+1)}$

Answer 1

$\frac{C_1}{4}suggest{\int }_0^1{C}_1{x}^{3}dx\frac{C_2}{7}$ suggest
${\int }_0^1{C}_2(x^3{)}^{2}dx,$ Hence consider ${\int }_2^1(1-x^3{)}^{n}dx$
${\int }_2^1[1-C_1{x}^3+C_2(x^3{)}^2+.....+(-{10}^n{C}_n(x^3{)}^n]dx$
$C_0-\frac{C_1}{4}+\frac{C_2}{7}-.......+(-1{)}^2\frac{C_n}{3n+1}$ = L. H. S.
Again $I_m={\int }_0^{1}1.(1-x^3{)}^{n}dx$
$=\left[x\right(1-x^3{)}^n]{\int }_0^1-n{\int }_0^{1}x(1-x^3{)}^n(-3x^2)dx$
We have integrated by parts
or $I_n=0+3n{\int }_0^1{x}^3(1-x^3{)}^{n-1}dx$
$=3n{\int }_0^1[1-(1-x^3\left)\right](1-x^3{)}^{n-1}$
$=3n[I_{n-1}-I_n]$
$\therefore I_n(3n+1)=3n{I}_{n-1}$
$\therefore I_n=\frac{3n}{}3n+1I_{n-1}=\frac{3n}{}3n+1\frac{3(n-1)}{3n-2}{I}_{n-1}$
$\frac{3n}{}3n+1\frac{3(n-1)}{3n-2}{I}_{n-1}.....\frac{3.1}{3.1+1}.I_0$
$=\frac{3^n.n!}{(3n+1)....7.4.1}1$ R. H. S.
as $I_0={\int }_0^1(1-x^3{)}^{0}dx={\int }_0^{1}1.dx=[x{]}_0^1=1$