∑ (from k=1 to n) (-1)^(k+1) 2k

Steps to Solve

1. Understanding the Summation Notation

The expression given is a summation: $$ \sum_{k=1}^{n} (-1)^{k+1} 2k $$ This means we will sum the terms generated by the formula $(-1)^{k+1} 2k$ for each integer value of $k$ starting from 1 and going up to $n$.

2. Evaluating the Terms of the Sum

Let's calculate a few terms to observe the pattern:

• For $k=1$: $(-1)^{1+1} 2(1) = 2$
• For $k=2$: $(-1)^{2+1} 2(2) = -4$
• For $k=3$: $(-1)^{3+1} 2(3) = 6$
• For $k=4$: $(-1)^{4+1} 2(4) = -8$

The series alternates between adding and subtracting values.

3. Formulating the Series

The series can be expressed as: $$ S = 2 - 4 + 6 - 8 + ... + (-1)^{n+1} 2n $$ We separate it into the sum of even and odd indexed elements.

4. Finding the General Formula

When $n$ is odd:

• Positive terms: $2 + 6 + ... + 2n = 2(1 + 3 + \ldots + n) \Rightarrow$ (Sum of first n odd numbers = $n^2$)
• Negative terms: $-4 - 8 - ... - 2(n-1) = -2(2 + 4 + ... + (n-1)) \Rightarrow$ (Arithmetic series sum)

When $n$ is even:

• Positive terms and negative terms can be summed similarly.

5. Expressing The Final Result

To find the final sum, we can derive that: For even $n$: $$ S_n = -n $$

For odd $n$: $$ S_n = n + 2 $$