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Question Number 107673 by qwerty111 last updated on 12/Aug/20

Answered by Her_Majesty last updated on 12/Aug/20

$$=\underset{k=1}{\sum ^n}(2^{2k}+2^{-2k}+2)=\underset{k=1}{\sum ^n}{4}^k+\underset{k=1}{\sum ^n}\frac{1}{4^k}+\underset{k=1}{\sum ^n}2=$$$$=\frac{4^{n+1}-4}{3}+\frac{4^n-1}{3\times 4^n}+2n=$$$$=\frac{4^{n+1}-4^{-n}}{3}+2n-1$$

Answered by Dwaipayan Shikari last updated on 12/Aug/20

$$(2^2+4^2+8^2+...+n)+(2+2+2+..+n)+(\frac{1}{4}+\frac{1}{16}+\frac{1}{64}+...n)$$$$4.\frac{4^n-1}{3}+2n+\frac{1}{4}.\frac{1-{\left(\frac{1}{4}\right)}^n}{1-\frac{1}{4}}$$$$4.\frac{4^n-1}{3}+2n+\frac{1-4^{-n}}{3}$$

Commented by Her_Majesty last updated on 12/Aug/20

$$yesnow{it}^{\prime }sthesameasmysolution$$

Answered by 1549442205PVT last updated on 12/Aug/20

$${(2^{\mathrm{n}}+\frac{1}{2^{\mathrm{n}}})}^2=4^{\mathrm{n}}+2+\frac{1}{4^{\mathrm{n}}}.\mathrm{Therefore},$$$$\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}}}{(2^{\mathrm{n}}+\frac{1}{2^{\mathrm{n}}})}^2=2\mathrm{n}+\underset{\mathrm{k}=1}{\stackrel{\mathrm{n}}{\mathrm{\Sigma }}}{4}^{\mathrm{n}}+\underset{\mathrm{k}=1}{\stackrel{\mathrm{n}}{\mathrm{\Sigma }}}\frac{1}{4^{\mathrm{n}}}$$$$=4.\frac{4^{\mathrm{n}}-1}{4-1}+\frac{1}{4}.\frac{1-{\left(\frac{1}{4}\right)}^{\mathrm{n}}}{1-\frac{1}{4}}=2\mathrm{n}+\frac{4}{3}(4^{\mathrm{n}}-1)$$$$+\frac{1}{3}[1-{\left(\frac{1}{4}\right)}^{\mathbf{n}}]=2\mathbf{n}-1+\frac{4^{\mathbf{n}+1}}{3}-\frac{1}{3.4^{\mathbf{n}}}$$

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