CMR:  $C^{0}_{n}+\frac{2}{2}C^{1}_{n}+\frac{2^{2}}{3}C^{2}_{n} + ... + \frac{2^{n-1}}{n}C^{n-1}_{n}+ \frac{2^{n}}{n+1}C^{n}_{n} =  \frac{3^{n+1}-1}{2(n+1)},\forall n \in N^{*}$
check lại để đi bạn –  Dép Lê Con Nhà Quê 20-12-13 12:01 PM

Câu hỏi này được treo giải thưởng trị giá +1000 vỏ sò bởi hohuynhnhutuyen@yahoo.com.vn, đã hết hạn vào lúc 20-12-13 05:05 PM

Ta có: $(2t+1)^n=\sum_{k=0}^{n}C^k_n(2t)^k$
$\Rightarrow (2t+1)^n=\sum_{k=0}^{n}2^kC^k_nt^k$
$\Rightarrow \int\limits_0^1(2t+1)^ndt=\int\limits_0^1\sum_{k=0}^{n}2^kC^k_nt^kdt$
$\Rightarrow \dfrac{(2t+1)^{n+1}}{2(n+1)}\left|\begin{array}{l}1\\0\end{array}\right.=\sum_{k=0}^n\dfrac{2^kC^k_nt^{k+1}}{k+1}\left|\begin{array}{l}1\\0\end{array}\right.$
$\Rightarrow \sum_{k=0}^n\dfrac{2^kC^k_n}{k+1}=\dfrac{3^{n+1}-1}{2(n+1)}$
mình mới hoc HKI lớp 11 ak –  hohuynhnhutuyen 21-12-13 12:14 PM
bạn có thể giải theo cach khác được không, cách lop 11, cái gạch xuống 1 0 mình không hiểu –  hohuynhnhutuyen 21-12-13 12:14 PM

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