Thu gọn \(2 \mathrm{C}_{n}^{0}+\frac{2^{2} \mathrm{C}_{n}^{1}}{2}+\frac{2^{3} \mathrm{C}_{n}^{2}}{3}+\frac{2^{4} \mathrm{C}_{n}^{3}}{4}+\cdots+\frac{2^{n+1} \mathrm{C}_{n}^{n}}{n+1}\) ta được
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Lời giải:
Báo saiTa có: \((1 x)^{n+1}=\mathrm{C}_{n+1}^{0}+\mathrm{C}_{n+1}^{1} x+\mathrm{C}_{n+1}^{2} x^{2}+\cdots+\mathrm{C}_{n+1}^{n+1} x^{n+1} .\)
\(\begin{array}{l} \text { Ta có: } \mathrm{C}_{n+1}^{k+1}=\frac{n+1}{k+1} \mathrm{C}_{n}^{k} \\ \text { Vậy }(1+x)^{n+1}=(n+1)\left(\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{0} x+\frac{1}{2} \mathrm{C}_{n}^{1} x^{2}+\frac{1}{3} \mathrm{C}_{n}^{2} x^{3}+\cdots+\frac{1}{n+1} \mathrm{C}_{n}^{0} x^{n+1}\right) . \\ \Leftrightarrow \frac{(1+x)^{n+1}}{n+1}=\mathrm{C}_{n}^{0}+\mathrm{C}_{n}^{0} x+\frac{1}{2} \mathrm{C}_{n}^{1} x^{2}+\frac{1}{3} \mathrm{C}_{n}^{2} x^{3}+\cdots+\frac{1}{n+1} \mathrm{C}_{n}^{n} x^{n-1} \end{array}\)
Cho x=2 ta được
\(\frac{3^{n+1}}{n+1}=1+2 \mathrm{C}_{n}^{0}+\frac{2^{2} \mathrm{C}_{n}^{1}}{2}+\frac{2^{3} \mathrm{C}_{n}^{2}}{3}+\frac{2^{4} \mathrm{C}_{n}^{3}}{4}+\cdots+\frac{2^{n-1} \mathrm{C}_{n}^{n}}{n+1}=\frac{3^{n+1}}{n+1}-1\)