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Question Number 115897 by Joel574 last updated on 29/Sep/20

$$\mathrm{Given}$$$$a_n=\sqrt{1+{(1-\frac{1}{n})}^2}+\sqrt{1+{(1+\frac{1}{n})}^2}$$$$\mathrm{The}\mathrm{value}\mathrm{of}\underset{n=1}{\sum ^{2015}}\left(\frac{4}{a_n}\right)\mathrm{is}...$$

Commented by Joel574 last updated on 29/Sep/20

$$\left(a\right)\sqrt{{2015}^2+{2016}^2}$$$$\left(b\right)\sqrt{{2015}^2+{2016}^2}+1$$$$\left(c\right)\sqrt{{2014}^2+{2015}^2}+1$$$$\left(d\right)\sqrt{{2015}^2+{2016}^2}-1$$$$\left(e\right)\sqrt{{2014}^2+{2015}^2}-1$$

Answered by mindispower last updated on 29/Sep/20

$$a_n=\frac{1}{n}(\sqrt{2n^2-2n+1}+\sqrt{2n^2+2n+1})$$$$\frac{1}{a_n}=\frac{n}{\sqrt{2n^2+2n+1}+\sqrt{2n^2-2n+1}}$$$$=\frac{n}{4n}(\sqrt{2n^2+2n+1}-\sqrt{2n^2-2n+1})$$$$=\frac{1}{4}(\sqrt{2n^2+2n+1}-\sqrt{2n^2-2n+1})$$$$u_n=\sqrt{2n^2-2\mathit{n}+1}$$$${\mathit{u}}_{\mathit{n}+1}=\sqrt{2{(\mathit{n}+1)}^2-2(\mathit{n}+1)+1}$$$$=\sqrt{2{\mathit{n}}^2+2\mathit{n}+1}$$$$\frac{1}{\mathit{a}\mathit{n}}=\frac{1}{4}({\mathit{U}}_{n+1}-U_n)$$$$\mathrm{\Sigma }\frac{1}{a_n}=\frac{1}{4}\mathrm{\Sigma }(u_{n+1}-u_n)$$$$\sum _{n⩽2015}\frac{4}{a_n}=U_{2016}-u_1$$$$u_1=1$$$$u_n=\sqrt{{(n+1)}^2+n^2}$$$$u_{2016}=\sqrt{{2017}^2+{2016}^2}$$$$\Rightarrow \mathrm{\Sigma }\frac{4}{a_n}=\sqrt{{2017}^2+{2016}^2}-1$$

Commented by Joel574 last updated on 30/Sep/20

$$Thankyouverymuch$$

Answered by Dwaipayan Shikari last updated on 29/Sep/20

$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}\left(\frac{4}{{\mathrm{a}}_{\mathrm{n}}}\right)=\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}\frac{4}{\sqrt{1+{(1-\frac{1}{\mathrm{n}})}^2}+\sqrt{1+{(1+\frac{1}{\mathrm{n}})}^2}}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}4\frac{\sqrt{1+{(1-\frac{1}{\mathrm{n}})}^2}-\sqrt{1+{(1+\frac{1}{\mathrm{n}})}^2}}{-\frac{4}{\mathrm{n}}}$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}\mathrm{n}(\sqrt{1+{(1+\frac{1}{\mathrm{n}})}^2}-\sqrt{1+{(1-\frac{1}{\mathrm{n}})}^2})$$$$\underset{\mathrm{n}=1}{\sum ^{\mathrm{n}}}\sqrt{{2\mathrm{n}}^2+2\mathrm{n}+1}-\sqrt{{2\mathrm{n}}^2-2\mathrm{n}+1}$$$$=\sqrt{5}-\sqrt{1}+\sqrt{13}-\sqrt{5}+...+\sqrt{{\mathrm{n}}^2+{(\mathrm{n}+1)}^2}-\sqrt{{(\mathrm{n}-1)}^2+{\mathrm{n}}^2}$$$$=\sqrt{{\mathrm{n}}^2+{(\mathrm{n}+1)}^2}-1$$$$$$

Commented by Joel574 last updated on 30/Sep/20

$$Thankyouverymuch$$

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