Câu hỏi của Lê Tùng Anh

Question

Tìm n ϵ N: 2.2222 + 3.2323 + 4.2424 +...+ n.2n2� = 2n+11

Akai Haruma · Accepted Answer

Lời giải:Đặt $A=2.2^2+3.2^3+4.2^4+...+n.2^n$$2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}$$\Rightarrow A=2A-A=n.2^{n+1}-(2^3+2^4+...+2^n) - 2.2^2$$\Rightarrow A=n.2^{n+1}-(2^3+2^4+...+2^n)-8$Đặt $S=2^3+2^4+...+2^n$$2S=2^4+2^5+...+2^{n+1}$$\Rightarrow S=2S-S=2^{n+1}-2^3=2^{n+1}-8$$\Rightarrow A=n.2^{n+1}-S-8 = n.2^{n+1}-2^{n+1}+8-8=(n-1).2^{n+1}$Vậy $(n-1).2^{n+1}=2^{n+11}$$\Rightarrow n-1 = 2^{10}\Rightarrow n=2^{10}+1=1025$Câu trả lời: Lời giải:Đặt $A=2.2^2+3.2^3+4.2^4+...+n.2^n$$2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}$$\Rightarrow A=2A-A=n.2^{n+1}-(2^3+2^4+...+2^n) - 2.2^2$$\Rightarrow A=n.2^{n+1}-(2^3+2^4+...+2^n)-8$Đặt $S=2^3+2^4+...+2^n$$2S=2^4+2^5+...+2^{n+1}$$\Rightarrow S=2S-S=2^{n+1}-2^3=2^{n+1}-8$$\Rightarrow A=n.2^{n+1}-S-8 = n.2^{n+1}-2^{n+1}+8-8=(n-1).2^{n+1}$Vậy $(n-1).2^{n+1}=2^{n+11}$$\Rightarrow n-1 = 2^{10}\Rightarrow n=2^{10}+1=1025$

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