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. $\frac{1}{2}(3^{40}-1)$
• D. $\frac{1}{2}(3^{40}+1)$

Answer 1

The correct option is D $\frac{1}{2}(3^{40}+1)$

$(1+2\sqrt{x}{)}^{40}={}^{40}{C}_0+{}^{40}{C}_1(2\sqrt{x})+{}^{40}{C}_2(2\sqrt{x}{)}^2+\cdots +{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}\cdots (1)$
The coefficients of integral powers of $x$ are
${}^{40}{C}_0,{}^{40}{C}_2\cdot 2^2,{}^{40}{C}_4\cdot 2^4,\dots ,{}^{40}{C}_{40}\cdot 2^{40}$

Now, $(1-2\sqrt{x}{)}^{40}$
$={}^{40}{C}_0-{}^{40}{C}_1\left(2\sqrt{x}\right)+{}^{40}{C}_2(2\sqrt{x}{)}^2+\cdots +{}^{40}{C}_{40}(2\sqrt{x}{)}^{40}\cdots \left(2\right)$
Adding equations $\left(1\right)$ and $\left(2\right)$, putting $x=1$, we get
$(3{)}^{40}+1=2[{}^{40}{C}_0+{}^{40}{C}_2\cdot 2^2+{}^{40}{C}_4\cdot 2^4+\cdots +{}^{40}{C}_{40}\cdot 2^{40}]$
$\therefore$ Required sum $=\frac{3^{40}+1}{2}$