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Câu 3
$1-\frac{1}{3}C_{n}^{1}+\frac{1}{5}C_{n}^{2}-\frac{1}{7}C_{n}^{3}+...+\frac{(-1)^n}{2n+1}C_{n}^{n}$.
Đây là một bài toán hay và khó! (Mất cả buổi chiều suy nghĩ
)
Ta viết lại đề:
$S_n=\sum_{k=0}^n \dfrac{(-1)^k C_n^k}{2k+1}$
Ta có:
$S_n=1+\sum_{k=1}^n \dfrac{(-1)^k C_n^k}{2k+1}=1+\sum_{k=1}^n\left[\dfrac{n}{n-k}.\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}\right]$
$S_n=1+\dfrac{1}{2n+1}\sum_{k=1}^n\left[\dfrac{n(2n+1)}{n-k}.\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}\right]$
$S_n=1+\dfrac{1}{2n+1}\sum_{k=1}^n\left[\dfrac{2n^2-2nk+2nk+n}{n-k}.\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}\right]$
$S_n=1+\dfrac{1}{2n+1}\sum_{k=1}^n\left[\dfrac{2n^2-2nk}{n-k}.\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}\right]+\dfrac{1}{2n+1}\sum_{k=1}^n\left[\dfrac{2nk+n}{n-k}.\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}\right]$
$S_n=1+\dfrac{2n}{2n+1}\sum_{k=1}^{n-1}\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}+\dfrac{1}{2n+1}\sum_{k=1}^n \dfrac{n(-1)^k C_{n-1}^k}{n-k}$
$S_n=\dfrac{2n}{2n+1}\sum_{k=0}^{n-1}\dfrac{(-1)^k C_{n-1}^k}{(2k+1)}+\dfrac{1}{2n+1}\sum_{k=0}^n \left[(-1)^k C_n^k\right]$
$\Rightarrow S_n=\dfrac{2n}{2n+1}S_{n-1}+0$
$\Rightarrow S_{n-1}=\dfrac{2n-2}{2n-1}S_{n-2}$
...
$\Rightarrow S_1=\dfrac{2}{3}S_0$
Dễ thấy $S_0=1$
$\Rightarrow S_n=\dfrac{2n(2n-2)...2}{(2n+1)(2n-1)...3.1}=\dfrac{(2n)!!}{(2n+1)!!}$
(Giai thừa cách đôi!
- Đừng nhầm với ! nhé!)