Question

In the usual notation prove that $2C_0+\frac{2^2}{2}{C}_1+\frac{2^3}{3}{C}_2+......+\frac{2^{11}}{11}{C}_{10}=\frac{3^{11}-1}{11}$

Answer 1

$(1+x{)}^n=C_0+C_{1}x+C_2{x}^2+C_3{x}^3+...C_n{x}^n$ ....(1)
We have to prove that
$2C_0+\frac{2^2}{2}{C}_1+\frac{2^3}{3}{C}_2+...+\frac{2^{11}}{11}{C}_{10}=\frac{3^{11}-1}{11}$
1 + x = 3 for x = 2. All the terms are +ive hence we integrate both sides of (1) with limits as 0 to 2
Integrating both sides of w.r.t x
${\left[\frac{(1+x{)}^{n+1}}{n+1}\right]}_0^2={[C_{0}x+C_1\cdot \frac{x^2}{2}+C_2\frac{x^3}{3}+....+C_n\frac{x^{n+1}}{n+1}]}_0^2$....(2)
Now putting x = 2 and n = 10 in (2) , we get the result