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2026-04-02 01:36:49.1775093809
Find the sum of coefficient of all the integral power of $x$ in the expansion of $\big(1 + 2\sqrt x\big)^{40}$?
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The answer is $\frac{1}{2} (3^{40} + 1)$. Let $$g(y) = (1 + 2 y)^{40}$$ Sought for is the sum of coefficients of $y^{2i}$ of the series expansion of $g(y)$. Now the subseries of $g(y)$ consisting only of even powers is obviously $h(y) = \frac{1}{2}(g(y) + g(-y))$. The sum of these coefficients is therefore $h(1) = \frac{1}{2}(g(1) + g(-1)) = \frac{1}{2}(3^{40} + (-1)^{40}) = \frac{1}{2}(3^{40} + 1)$