Arithmetic Series Question - Edexcel C2

Question

I'm trying to solve the the following series

$$\sum_{i=1}^{2k} (-1)^{n+1}n $$

The answer to the series is $$-k$$

This is what I've tried:

$$\sum_{i=1}^{2k} (-1)^{n+1}n =(-1)^{1+1}*1+(-1)^{2k+1}*2k$$

This is not the correct answer according to the book.

I'm struggling to find the proof. Can someone help?

Answer 1

Edit: After reading many of the comments, it appears as though there is a misprint in your text. I believe the solution I provided is addressing the intended question.

First, make sure you understand the sum notation: $$\sum_{i=1}^{2k}(-1)^{i+1}i = (-1)^{1+1} + (-1)^{2+1} +\cdots+(-1)^{2i+1}2i.$$ Notice that the terms in this sum alternate between positive and negative values. This means there is going to be some cancellation. We may rewrite the sum as $$ \sum_{i=1}^k(-1)^{2i}(2i-1) + (-1)^{2i+1}(2i). $$ Here we are just adding two terms at a time from the first sum. Notice that each of these new terms is in fact equal to $-1$. Thus, $$\sum_{i=1}^{2k}(-1)^{i+1}i = -k$$

Answer 2

Consider writing out the first few terms:

$$S=1-2+3-4+5-6...$$

and you should see a pattern:

$$S=(1-2)+(3-4)+(5-6)...$$

Can you tell how many such pairs are there?

Answer 3

Write it as $$ \sum_{n=1}^{2k}(-1)^{n+1}n=\sum_{n=1}^k((-1)^{2n-1+1}(2n-1)+(-1)^{2n+1}(2n))=\sum_{n=1}^{k}(-1)=-k. $$

Answer 4

You started well. Note that you plug all numbers from $1$ to $2k$ when you expand summation, not only these two. Here is the full expansion (and note the red index must be a misprint in the book): $$\sum_{\color{red}{n}=1}^{2k} (-1)^{n+1}n=(-1)^{1+1}\cdot 1+(-1)^{2+1}\cdot 2+(-1)^{3+1}\cdot 3+(-1)^{4+1}\cdot 4+\cdots +(-1)^{2k-1+1}\cdot (2k-1)+(-1)^{2k+1}\cdot 2k=$$ $$1-2+3-4+\cdots +(2k-1)-2k=\underbrace{-1-1-\cdots -1}_{k}=-k.$$