Question

The sum of the coefficients of all the integral powers of $x$ in the expansion of $(1+2\sqrt{x}{)}^{40}$ is :

• A. $3^{40}+1$
• B. $3^{40}-1$
• C. $\frac{1}{2}\left(3^{40}-1\right)$
• D. $\frac{1}{2}\left(3^{40}+1\right)$

Answer 1

Answer: (D)

$(1+2\sqrt{x}{)}^{40}={}^{40}{C}_0+{}^{40}{C}_{1}2\sqrt{x}+\dots ..+{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}$
$(1-2\sqrt{x}{)}^{40}={}^{40}{C}_0-{}^{40}{C}_{1}2\sqrt{x}+\dots \dots +{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}$
$(1+2\sqrt{x}{)}^{40}+(1-2\sqrt{x}{)}^{40}$
$=2\left[{}^{40}{C}_0+{}^{40}{C}_2(2\sqrt{x}{)}^2+\dots \dots +{}^{40}{C}_{40}\right](2\sqrt{x}{)}^{40}$
Putting $x=1$
${}^{40}{C}_0+{}^{40}{C}_2(2{)}^2+\dots \dots ..+{}^{40}{C}_{40}(2{)}^{40}=\frac{3^{40}+1}{2}$