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(A) Let $f(x) = (x+y)^{100} + (x-y)^{100}$.Using the binomial expansion,$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ and $(x-y)^{n} = \sum_{k

Vedclass · Accepted Answer

$51$

Vedclass · Answer

(A) Let $f(x) = (x+y)^{100} + (x-y)^{100}$.Using the binomial expansion,$(x+y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} y^k$ and $(x-y)^{n} = \sum_{k=0}^{n} \binom{n}{k} x^{n-k} (-y)^k$.Adding these,the odd terms cancel out,leaving only the even terms: $2 \sum_{k=0, 2, 4, \dots, n} \binom{n}{k} x^{n-k} y^k$.For $n = 100$,the number of terms is $\frac{n}{2} + 1 = \frac{100}{2} + 1 = 51$.

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

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