Chủ đề này có 1 trả lời

[faraanh]

tính $100C_{100}^{1}(\frac{1}{2})^{99}-101C_{100}^{1}(\frac{1}{2})^{100}+...-199C_{100}^{99}(\frac{1}{2})^{198}+200C_{100}^{100}(\frac{1}{2})^{199}$

[nthoangcute]

tính $100C_{100}^{1}(\frac{1}{2})^{99}-101C_{100}^{1}(\frac{1}{2})^{100}+...-199C_{100}^{99}(\frac{1}{2})^{198}+200C_{100}^{100}(\frac{1}{2})^{199}$

 

 

Tổng quát luôn cho máu:

$$S=\sum^n_{{k=1}} \dfrac{(n+k) (-1)^{k+1}}{2^{n+k-1}} \binom{n}{k}$$

Ta có:

$$\,{\frac {2 \left( k+1 \right)  \left( -1 \right) ^{k+1}}{{2}^{n+k}}}{n\choose k+1}-\,{\frac {2k \left( -1 \right) ^{k}}{{2}^{n+k-1}}}{n\choose k}\\=\frac{(k+1)(-1)^{k+1}}{2^{n+k-1}} \frac{n-k}{k+1} \binom{n}{k} - \,{\frac {2k \left( -1 \right) ^{k}}{{2}^{n+k-1}}}{n\choose k}\\=\frac{(-1)^{k+1} (n-k)}{2^{n+k-1}} \binom{n}{k} - \,{\frac {2k \left( -1 \right) ^{k}}{{2}^{n+k-1}}}{n\choose k}\\=\frac{(n+k) (-1)^{k+1}}{2^{n+k-1}} \binom{n}{k}$$

 

Suy ra:

$$\sum _{{k=1}}^n \frac{(n+k) (-1)^{k+1}}{2^{n+k-1}} \binom{n}{k}\\= \sum^n_{k=1} \left ( \,{\frac {2 \left( k+1 \right)  \left( -1 \right)^{k+1}}{{2}^{n+k}}}{n\choose k+1}-\,{\frac {2k \left( -1 \right) ^{k}}{{2}^{n+k-1}}}{n\choose k} \right )\\=\,{\frac {2 \left( n+1 \right)  \left( -1 \right) ^{n+1}}{{2}^{n+n}}}{n\choose n+1}- \,{\frac {2 \left( -1 \right) }{{2}^{n+1-1}}}{n\choose 1}\\=\frac{n}{2^{n-1}}$$