Calculate sum of ((2k-1)/2^k) for k = 1 to n

We are asked to compute the sum of the series:

\[
S_n = \sum_{k=1}^{n} \frac{2k-1}{2^k}
\]

Let's break this down:

\[
S_n = \sum_{k=1}^{n} \left( \frac{2k}{2^k} - \frac{1}{2^k} \right)
\]

This separates into two sums:

\[
S_n = 2 \sum_{k=1}^{n} \frac{k}{2^k} - \sum_{k=1}^{n} \frac{1}{2^k}
\]

### First sum: \( \sum_{k=1}^{n} \frac{1}{2^k} \)
This is a geometric series:

\[
\sum_{k=1}^{n} \frac{1}{2^k} = 1 - \frac{1}{2^n}
\]

For large \( n \), the sum approaches 1.

### Second sum: \( \sum_{k=1}^{n} \frac{k}{2^k} \)

This is a known sum that can be derived using a generating function:

\[
\sum_{k=1}^{\infty} \frac{k}{2^k} = \frac{2}{(1-1/2)^2} = 2
\]

However, for a finite sum, we have the following closed form:

\[
\sum_{k=1}^{n} \frac{k}{2^k} = 2 - \frac{n+2}{2^n}
\]

### Putting it all together

Substitute both sums back into the original expression:

\[
S_n = 2 \left( 2 - \frac{n+2}{2^n} \right) - \left( 1 - \frac{1}{2^n} \right)
\]

Simplifying this expression:

\[
S_n = 4 - \frac{2(n+2)}{2^n} - 1 + \frac{1}{2^n}
\]

\[
S_n = 3 - \frac{2n + 3}{2^n}
\]

Thus, the sum of the series is:

\[
S_n = 3 - \frac{2n + 3}{2^n}
\]

### Would you like any further details or clarifications?

Here are some related questions to explore:
1. How does this sum behave as \( n \to \infty \)?
2. Can this method be applied to different series of similar forms?
3. What are the practical applications of series like this one?
4. How would the sum change if the denominator were \( 3^k \) instead of \( 2^k \)?
5. Can we derive this sum using a different method, such as integration?

**Tip:** Understanding generating functions can be extremely helpful when working with sums involving powers of \( k \).

Explore the step-by-step solution to calculating the sum of the series ((2k-1)/2^k) from k = 1 to n. This problem covers advanced mathematical concepts including series summation, geometric series, and generating functions, suitable for Grades 11-12 and undergraduate students.

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