\(\text { Tính } \lim u_{n} \text { với } u_{n}=\frac{2.2^{2}+3.2^{3}+\ldots+n \cdot 2^{n}}{(n-1)\left(2^{n}+1\right)} \text { . }\)
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Lời giải:
Báo sai\(\begin{aligned} &\text { Đặt } S_{n}=2.2^{2}+3.2^{3}+\ldots+n \cdot 2^{n} \\ &\text { Khi đó } S_{n}+2=2+2.2^{2}+3.2^{3}+4.2^{4}+5.2^{5}+\ldots+n \cdot 2^{n} \\ &=\left(2+2^{2}+\ldots+2^{n}\right)+\left(2^{2}+2^{3}+\ldots+2^{n}\right)+\ldots+\left(2^{n-1}+2^{n}\right)+2^{n} \\ &=\frac{2\left(1-2^{n}\right)}{1-2}+\frac{2^{2}\left(1-2^{n-1}\right)}{1-2}+\ldots+\frac{2^{n-1}\left(1-2^{2}\right)}{1-2}+\frac{2^{n}\left(1-2^{1}\right)}{1-2} \\ &=n .2^{n+1}-\left(2+2^{2}+\ldots+2^{n}\right)=n \cdot 2^{n+1}-\frac{2\left(1-2^{n}\right)}{1-2}=(n-1) \cdot 2^{n+1}+2 \\ &\text { Suy ra } S_{n}+2=(n-1) \cdot 2^{n+1}+2 \Leftrightarrow S_{n}=(n-1) \cdot 2^{n+1} \\ &\text { Vậy } \lim u_{n}=\lim \frac{S_{n}}{(n-1)\left(2^{n}+1\right)}=\lim \frac{(n-1) \cdot 2^{n+1}}{(n-1)\left(2^{n}+1\right)}=\lim \frac{2^{n+1}}{2^{n}+1}=\lim \frac{2}{1+\left(\frac{1}{2}\right)^{n}}=2 \end{aligned}\)
