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Mathematics, 02.08.2019 06:00 cheating53

For each of 2 ≤ n ≤ 5, compute c(n, 0) + c(n, 2) + ⋅⋅⋅ + c(n, 2k), where 2k is largest even integer ≤ n, and compute c(n, 1) + c(n, 3) + ⋅⋅⋅ + c(n, 2k + 1), where 2k + 1 is the largest odd integer ≤ n.

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For each of 2 ≤ n ≤ 5, compute c(n, 0) + c(n, 2) + ⋅⋅⋅ + c(n, 2k), where 2k is largest even integer...
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