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Solution

$(x+a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}a^r$ $(x-a)^{100} = \sum_{r=0}^{100} \binom{100}{r} x^{100-r}(-a)^r$ Adding, odd powers of $a$ cancel and even powers remain. Even $r$: total even $r$ from 0 to 100 → 51 terms.