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

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\frac{3^{40} - 1}{2}

\frac{3^{40} + 1}{2}

\frac{3^{38} - 1}{2}

\frac{3^{38} + 1}{2}

answer is B.

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Detailed Solution

{(1 + 2 \sqrt{x})}^{40}

The coefficient of the integral power of $x$ are

$^{40} C_{0} .^{40} C_{2} {.2}^{2} .^{40} C_{4} 2^{4} \dots^{40} C_{40} 2^{40}$

${(1 + 2)}^{40} =^{40} C_{0} +^{40} C_{1} {.2}^{1} +^{40} C_{2} {.2}^{2} + \dots +^{40} C_{40} 2^{40} \dots \dots (1)$

$\underline{{(1 - 2)}^{40} =^{40} C_{0} -^{40} C_{1} {.2}^{1} +^{40} C_{2} {.2}^{2} + \dots +^{40} C_{40} 2^{40} \dots \dots (2)}$

$3^{40} + 1 = 2 [^{40} C_{0} +^{40} C_{2} {.2}^{2} + \dots +^{40} C_{40} 2^{40}]$

$^{40} C_{0} +^{40} C_{2} {.2}^{2} + \dots +^{40} C_{40} 2^{40} = \frac{3^{40} + 1}{2}$

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