Question

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.

Answer 1

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)`