Question

For the expansion ${(1+2\sqrt{x})}^{40}$, sum of the coefficients of the

• A. integral powers of $x$ is $\frac{(3^{40}+1)}{2}$
• B. non-integral powers of $x$ is $\frac{(3^{40}-1)}{2}$
• C. integral powers of $x$ is $\frac{(3^{40}-1)}{2}$
• D. non-integral powers of $x$ is $\frac{(3^{40}+1)}{2}$

Answer 1

Answer: (A), (B)

integral powers of $x$ is $\frac{(3^{40}+1)}{2}$
non-integral powers of $x$ is $\frac{(3^{40}-1)}{2}$

The integral terms will be the odd number terms.
$={}^{40}{C}_0+{}^{40}{C}_{2}2+...$
$=\frac{1}{2}\left[\right(1+2{)}^{40}+(1-2{)}^{40}]$
$=\frac{3^{40}+1}{2}$
Now total sum of coefficients is $3^{40}$.
Hence sum of the coefficients of non-integral terms will be
$=3^{40}-\frac{3^{40}+1}{2}$
$=\frac{3^{40}-1}{2}$.