Chủ đề này có 1 trả lời

[nthoangcute]

Tình tổng sau:
$$S=\sum^n_{k=0} 3^{2k}\binom{2n+1}{2k}$$

[dark templar]

Tình tổng sau:
$$S=\sum^n_{k=0} 3^{2k}\binom{2n+1}{2k}$$

Xét khai triển sau $(x+3)^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}3^{k}x^{2n+1-k}$

 

Cho $x=1$ thì $\sum_{k=0}^{2n+1}\binom{2n+1}{k}3^{k}=4^{2n+1}$

 

Cho $x=-1$ thì $2^{2n+1}=\sum_{k=0}^{2n+1}\binom{2n+1}{k}3^{k}(-1)^{2n+1-k}=\sum_{k=0}^{2n+1}(-1)^{k+1}3^{k}\binom{2n+1}{k}$.

 

Suy ra :

$$\sum_{k=0}^{n}3^{2k}\binom{2n+1}{2k} = \frac{{\sum\limits_{k = 0}^{2n + 1} {{3^k}\dbinom{2n + 1}{k}}  - \sum\limits_{k = 0}^{2n + 1} {{3^k}{{\left( { - 1} \right)}^{k + 1}}\dbinom{2n + 1}{k}} }}{2} = \frac{{{4^{2n + 1}} - {2^{2n + 1}}}}{2} = {2^{4n + 1}} - {2^{2n}}$$