MHB Find The Sum ∑(1/[3^n+√(3^(2017)]

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The discussion focuses on evaluating the sum ∑(1/[3^n+√(3^(2017))]) from n=0 to 2017. Participants express appreciation for the solutions provided, particularly highlighting a contribution from a user named Sudharaka. The evaluation involves simplifying the expression and understanding the behavior of the terms as n varies. The conversation emphasizes the mathematical techniques used to arrive at the solution. Overall, the thread showcases collaborative problem-solving in evaluating a complex mathematical sum.
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Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$
 
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lfdahl said:
Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$
$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}{3^{2015}+\sqrt{3^{2017}}}+\frac{1}{3^{2016}+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}$$

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,

$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\left(\frac{2018}{2}\right)\frac{1}{\sqrt{3^{2017}}}=\frac{1009}{\sqrt{3^{2017}}}$$
 
Sudharaka said:
$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\frac{1}{1+\sqrt{3^{2017}}}+\frac{1}{3+\sqrt{3^{2017}}}+\frac{1}{3^2+\sqrt{3^{2017}}}+\cdots+\frac{1}{3^{2015}+\sqrt{3^{2017}}}+\frac{1}{3^{2016}+\sqrt{3^{2017}}}+\frac{1}{3^{2017}+\sqrt{3^{2017}}}$$

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
 

Thread 'Area of Overlapping Squares'

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