MHB Find The Sum ∑(1/[3^n+√(3^(2017)]

[lfdahl]

Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$

 

[Sudharaka]

Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$

Spoiler

$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}{3^{2015}+\sqrt{3^{2017}}}+\frac{1}{3^{2016}+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}$$

Observe that taking the first and last terms;

$$\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{\sqrt{3^{2017}}(1+\sqrt{3^{2017}})}=\frac{1}{\sqrt{3^{2017}}}$$

Generally this holds for all $i$ and $n-i$ terms taken in pairs. Thus our sum reduces to,

$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\left(\frac{2018}{2}\right)\frac{1}{\sqrt{3^{2017}}}=\frac{1009}{\sqrt{3^{2017}}}$$

 

[lfdahl]

Spoiler

$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}{3^{2015}+\sqrt{3^{2017}}}+\frac{1}{3^{2016}+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}$$

Observe that taking the first and last terms;

$$\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{\sqrt{3^{2017}}(1+\sqrt{3^{2017}})}=\frac{1}{\sqrt{3^{2017}}}$$

Generally this holds for all $i$ and $n-i$ terms taken in pairs. Thus our sum reduces to,

$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\left(\frac{2018}{2}\right)\frac{1}{\sqrt{3^{2017}}}=\frac{1009}{\sqrt{3^{2017}}}$$

What a nice solution, Sudharaka! Thankyou very much for your participation