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(B) The binomial expansion of $(x + a)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.For $(x + a)^{100} + (x - a)^{100}$,the terms with odd

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$51$

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(B) The binomial expansion of $(x + a)^n$ is given by $\sum_{k=0}^{n} \binom{n}{k} x^{n-k} a^k$.For $(x + a)^{100} + (x - a)^{100}$,the terms with odd powers of $a$ cancel out.Specifically,$(x + a)^{100} = \binom{100}{0}x^{100} + \binom{100}{1}x^{99}a + \binom{100}{2}x^{98}a^2 + \dots + \binom{100}{100}a^{100}$.And $(x - a)^{100} = \binom{100}{0}x^{100} - \binom{100}{1}x^{99}a + \binom{100}{2}x^{98}a^2 - \dots + \binom{100}{100}a^{100}$.Adding these two expressions,the odd terms cancel out,leaving only the even terms: $2[\binom{100}{0}x^{100} + \binom{100}{2}x^{98}a^2 + \dots + \binom{100}{100}a^{100}]$.The indices of the remaining terms are $k = 0, 2, 4, \dots, 100$.The number of such terms is $\frac{100 - 0}{2} + 1 = 50 + 1 = 51$.

The total number of terms in the expansion of $(x + a)^{100} + (x - a)^{100}$ after simplification will be

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