The sum of the coefficients of the integral p | vedclass

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(B) Let $f(x) = (1 - 2\sqrt{x})^{50} = \sum_{r=0}^{50} {^{50}C_r} (-2\sqrt{x})^r$.The general term is $T_{r+1} = {^{50}C_r} (-2)^r x^{r/2}$.For the po

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$\frac{1}{2}(3^{50} + 1)$

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(B) Let $f(x) = (1 - 2\sqrt{x})^{50} = \sum_{r=0}^{50} {^{50}C_r} (-2\sqrt{x})^r$.The general term is $T_{r+1} = {^{50}C_r} (-2)^r x^{r/2}$.For the power of $x$ to be an integer,$r$ must be an even integer. Let $r = 2k$,where $k \in \{0, 1, 2, \dots, 25\}$.The terms with integral powers of $x$ are $\sum_{k=0}^{25} {^{50}C_{2k}} (-2)^{2k} x^k = \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k} x^k$.The sum of these coefficients is $S = \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k}$.Consider the expansions of $(1+2)^{50}$ and $(1-2)^{50}$:$(1+2)^{50} = \sum_{r=0}^{50} {^{50}C_r} 2^r = {^{50}C_0} + {^{50}C_1} 2^1 + {^{50}C_2} 2^2 + \dots + {^{50}C_{50}} 2^{50}$$(1-2)^{50} = \sum_{r=0}^{50} {^{50}C_r} (-2)^r = {^{50}C_0} - {^{50}C_1} 2^1 + {^{50}C_2} 2^2 - \dots + {^{50}C_{50}} 2^{50}$Adding these two equations:$(1+2)^{50} + (1-2)^{50} = 2 \sum_{k=0}^{25} {^{50}C_{2k}} 2^{2k}$.$3^{50} + (-1)^{50} = 2S$.$3^{50} + 1 = 2S$.$S = \frac{3^{50} + 1}{2}$.

The sum of the coefficients of the integral powers of $x$ in the binomial expansion of $(1 - 2\sqrt{x})^{50}$ is:

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