https://file.aiquickdraw.com/mathbot/file/17257369722533mgm3e3y.jpg

The mathematical expression given in the image is:

\[
\sum_{n=1}^{\infty} \frac{2n}{3^n}
\]

This is an infinite series, and to solve it, we can break it down step by step.

### General Approach:

We are summing terms of the form \( \frac{2n}{3^n} \), where \( n \) ranges from 1 to infinity.

This series can be handled using a standard summation technique for series of the form \( \sum_{n=1}^{\infty} n x^n \), where \( x \) is a constant ratio. The formula for this kind of series is:

\[
\sum_{n=1}^{\infty} n x^n = \frac{x}{(1 - x)^2} \quad \text{for} \quad |x| < 1
\]

In our case, we have \( x = \frac{1}{3} \), and we can adjust the formula to account for the factor \( 2 \).

Let me calculate the result step by step.The value of the infinite series 

\[
\sum_{n=1}^{\infty} \frac{2n}{3^n}
\]

is \( \frac{3}{2} \).

Would you like more details on how this result was derived, or do you have any other questions?

Here are 5 related questions to explore further:
1. How would the result change if the denominator were \( 4^n \) instead of \( 3^n \)?
2. What is the general formula for sums involving \( n \) in the numerator?
3. Can this type of sum be applied to geometric progressions with different starting indices?
4. What happens if we add a constant term to the summand, such as \( \frac{2n + 1}{3^n} \)?
5. How can you determine the convergence of an infinite series in general?

**Tip**: When dealing with infinite series, always check the convergence criteria, especially if the series involves ratios or powers.

Calculate the infinite series: sum from n=1 to infinity of 2n divided by 3^n.

Learn how to calculate the infinite series sum of 2n divided by 3^n using the geometric series formula. This detailed solution covers step-by-step calculations and the application of summation formulas, making it ideal for undergraduate-level students.

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