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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 ≤ n, and compute c(n, 1) + c(n, 3) + ⋅⋅⋅ + c(n, 2k + 1), where 2k + 1 is the largest odd integer ≤ n.

Respuesta :

LammettHash
Denote [tex]C(n,m)=\dbinom nm[/tex]. Then

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

In general, it would appear that

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

On the other hand,

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

so that in general, we also get

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

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