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ddd [48]
3 years ago
13

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.
Mathematics
1 answer:
Ksivusya [100]3 years ago
7 0
Denote C(n,m)=\dbinom nm. Then

n=2\implies\displaystyle\binom20+\binom22=1+1=2
n=3\implies\displaystyle\binom30+\binom32=1+3=4
n=4\implies\displaystyle\binom40+\binom42+\binom44=1+6+1=8
n=5\implies\displaystyle\binom50+\binom52+\binom54=1+10+5=16

In general, it would appear that

\displaystyle\sum_{k=0}^{2\lfloor\frac n2\rfloor}\binom n{2k}=2^{n-1}

On the other hand,

n=2\implies\displaystyle\binom21=2
n=3\implies\displaystyle\binom31+\binom33=3+1=4
n=4\implies\displaystyle\binom41+\binom43=4+4=8
n=5\implies\displaystyle\binom51+\binom53+\binom55=5+10+1=16

so that in general, we also get

\displaystyle\sum_{k=0}^{2\lfloor\frac{n-1}2\rfloor+1}\binom n{2k+1}=2^{n-1}
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