Question

Prove that $\frac{C_0}{1}+\frac{C_2}{3}+\frac{C_4}{5}+\frac{C_6}{7}+......=\frac{2^n}{n+1}$

Answer 1

$(1+x{)}^n+(1-x{)}^n=2[C_0+C_2{x}^2+C_4{x}^4+...]$
$(1+x{)}^n-(1-x{)}^n=2[C_{1}x+C_3{x}^3+C_5{x}^5+...]$
Now integrate both sides within limits 0 to 1 to get the desired results as given
${[\frac{(1+x{)}^{n+1}}{n+1}\pm \frac{(1-x{)}^{n+1}}{n+1}]}_0^1=2{[C_{0}x+C_2\frac{x^3}{3}+C_4\frac{x^5}{5}+...]}_0^1$
or $=2{[C_1\frac{x^2}{2}+C_3\frac{x^4}{4}+C_4\frac{x^6}{6}+...]}_0^1$
or $\frac{1}{2(n+1)}\left[\right(2^{n+1}-1)\pm (0-1\left)\right]=(\frac{16\left(a\right)}{16\left(b\right)})$
or $\frac{n^2}{n+1}=16$
for Q 16, 17, 18 , 19.see method of
$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+....C_n{x}^n$
Integrating w.r.t.x
$\frac{(1+x{)}^{n+1}}{n+1}l=C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}+....C_n\frac{x^{n+1}}{n+1}+A$
Where is a constant of integration and putting x = 0 in both sides, we get
$A=\frac{1}{n+1}$
$\therefore \frac{(1+x{)}^{n+1}}{n+1}-\frac{1}{n+1}$
$C_{0}x+C_1\frac{x^2}{2}+C_2\frac{x^3}{3}$ + ....(1)
Now putting x = 1 and - 1 in above , we get
$\frac{2^{n+1}-1}{n+1}=C_0+\frac{C_1}{2}+\frac{C_2}{3}+....\frac{C_n}{n+1}$ ;(x = 1)...(2)