Question

Find the sum of coefficients of odd powers of $x$ in the expansion ${(1+x)}^{50}$

Answer 1

${(1+x)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{+}^n{C}_0{x}^n$ ....$\left(1\right)$

${(1-x)}^n{=}^n{C}_0{-}^n{C}_{1}x{+}^n{C}_2{x}^2-...{+}^n{C}_0{x}^n$ ....$\left(2\right)$

$\Rightarrow {(1+x)}^n-{(1-x)}^n{=}^n{C}_0{+}^n{C}_{1}x{+}^n{C}_2{x}^2+...{+}^n{C}_0{x}^n{-}^n{C}_0{+}^n{C}_{1}x{-}^n{C}_2{x}^2+...{+}^n{C}_0{x}^n$

$=2({}^n{C}_{1}x{+}^n{C}_3{x}^3+....+x^n)$

Given: $n=50$ and put $x=1$

$\Rightarrow {(1+1)}^{50}-{(1-1)}^{50}=2^{50}$

$\Rightarrow 2^{50}=2({}^n{C}_{1}x{+}^n{C}_3{x}^3+....+x^n)$

$\therefore$Sum of odd powers of $x$ in the expansion is ${=}^n{C}_{1}x{+}^n{C}_3{x}^3+....+x^n=2^{50-1}=2^{49}$