The question asks to find the sum of the infinite series given by:

$\sum_{n \geq 0} \frac{n^2}{2^n}$

This is an infinite series where each term is of the form $\frac{n^2}{2^n}$. Let's solve it step by step.

Step 1: Recognize the series structure

The series is a power series with a general term $\frac{n^2}{2^n}$. Such series can often be computed using known formulas or through differentiation of simpler geometric series.

Step 2: Utilize known series result

One way to handle series involving $n^2$ is by using the fact that:

$\sum_{n=0}^{\infty} \frac{n^2}{x^n} = \frac{x(x+1)}{(x-1)^3}$

By substituting $x = 2$ (since our series involves powers of 2), we get:

$\sum_{n=0}^{\infty} \frac{n^2}{2^n} = \frac{2(2+1)}{(2-1)^3} = \frac{6}{1^3} = 6$

Final Answer:

The sum of the series is 6, which corresponds to option B.

Would you like more details on the steps? Here are five related questions you might find helpful:

1. How can we derive the sum formula for series involving $n^2$?
2. What is a geometric series, and how is it related to power series?
3. How does the differentiation of a geometric series help in solving power series involving $n^2$?
4. Are there other methods to compute such series without using standard results?
5. Can the same approach be used for sums involving higher powers of $n$ (e.g., $n^3$)?

Tip: When dealing with series, recognizing patterns in summation formulas can significantly speed up the calculation!