Cho $n$ là các số tự nhiên. Tính: $T=C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}$.

Q: Cho $n$ là các số tự nhiên. Tính: $T=C_{n}^{0}+\frac{1}{2} C_{n}^{1}+\frac{1}{3} C_{n}^{2}+\ldots+\frac{1}{n+1} C_{n}^{n}$.

Vì $C_{n}^{0}=1=\frac{1}{n+1} C_{n+1}^{1}$ và áp dụng công thức $\frac{1}{k+2} C_{n+1}^{k+1}=\frac{1}{n+2} C_{n+2}^{k+2}$ ta có:$\begin{aligned} T & =\frac{1}{n+1} C_{n+1}^{1}+\frac{1}{n+1} C_{n+1}^{2}+\frac{1}{n+1} C_{n+1}^{3}+\ldots+\frac{1}{n+1} C_{n+1}^{n+1} \\ & =\frac{1}{n+1}\left(C_{n+1}^{1}+C_{n+1}^{2}+C_{n+1}^{3}+\ldots+C_{n+1}^{n+1}\right)=\frac{1}{n+1}\left(2^{n+1}-1\right) .\end{aligned}$

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Accepted Answer

Vì $C_{n}^{0}=1=\frac{1}{n+1} C_{n+1}^{1}$ và áp dụng công thức $\frac{1}{k+2} C_{n+1}^{k+1}=\frac{1}{n+2} C_{n+2}^{k+2}$ ta có:

$\begin{aligned} T & =\frac{1}{n+1} C_{n+1}^{1}+\frac{1}{n+1} C_{n+1}^{2}+\frac{1}{n+1} C_{n+1}^{3}+\ldots+\frac{1}{n+1} C_{n+1}^{n+1} \ & =\frac{1}{n+1}\left(C_{n+1}^{1}+C_{n+1}^{2}+C_{n+1}^{3}+\ldots+C_{n+1}^{n+1}\right)=\frac{1}{n+1}\left(2^{n+1}-1\right) .\end{aligned}$

Đề thi cuối kì 2 (Cấu trúc mới) - CD - Đề số 1 - MĐ 9790