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Q.
The sum of the coefficients of all the integral powers of $x$ in the expansion of $(1+2 \sqrt{x})^{40}$ is :
Binomial Theorem
Solution:
$(1+2 \sqrt{x})^{40}={ }^{40} C _0+{ }^{40} C _1 2 \sqrt{ x }+\ldots . .+{ }^{40} C _{40}(2 \sqrt{ x })^{40}$
$(1-2 \sqrt{ x })^{40}={ }^{40} C _0-{ }^{40} C _1 2 \sqrt{ x }+\ldots \ldots+{ }^{40} C _{40}(2 \sqrt{ x })^{40}$
$ (1+2 \sqrt{x})^{40}+(1-2 \sqrt{x})^{40} $
$=2\left[{ }^{40} C _0+{ }^{40} C _2(2 \sqrt{ x })^2+\ldots \ldots+{ }^{40} C _{40}\right](2 \sqrt{ x })^{40}$
Putting $x=1$
${ }^{40} C _0+{ }^{40} C _2(2)^2+\ldots \ldots . .+{ }^{40} C _{40}(2)^{40}=\frac{3^{40}+1}{2}$