lfdahl said:Evaluate the sum:$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}$$
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,
$$\sum_{n=0}^{2017}\frac{1}{3^n+\sqrt{3^{2017}}}=\left(\frac{2018}{2}\right)\frac{1}{\sqrt{3^{2017}}}=\frac{1009}{\sqrt{3^{2017}}}$$
The number 2017 is the exponent of 3 in the denominator of the equation. It was likely chosen because it is a prime number, making the calculation more complex.
The value of n determines the number of terms in the sum. As n increases, the number of terms also increases, resulting in a larger sum.
Yes, the equation can be simplified by using the formula for the sum of a geometric series. This will result in a final value that is a function of n.
The value of √(32017) does not affect the sum as it is a constant value. It simply adds a small amount to the denominator of each term, making the overall sum slightly smaller.
This equation may be used in various fields such as mathematics, physics, and computer science to study the behavior of geometric series and their applications. It may also be used to model real-world phenomena that follow a geometric pattern.