Summation of an alternating series of squares?

I'm not sure what "something that may be broadly applicable" means. Every problem is a little different.

Here, the sum is the same as the sum of the integers from 1 to n, but with alternating sign.

1 1 1

2 -4 -3

3 9 6

4 -16 -10

5 25 15

6 -36 -21

7 49 28

8 -64 -36

9 81 45

10 -100 -55

11 121 66

So you get (-1)^(n+1) times n(n+1)/2

Your result can also be expressed as Sum(n^2) = n(n + 1)(n + 2)/6

Now consider your alternating series to an EVEN number 2n terms.

We expect the result to be negative, because the even terms are larger than the odd ones preceding.

Consider the example with the total series having 6 terms, (n = 3)

Let S1 = 1^2 + 3^2 + 5^2.....

We replace n by (2n - 1) in your formula

Sum(2n - 1)^2 = (2n - 1)2n(2n + 1)/6

Testing that out on the 3 terms above

{1 to 3}Sum(2n - 1)^2 = 5*6*7/6 = 35

1 + 9 + 25 = 35 so that was OK

Now let S2 = 2^2 + 4^2 + 6^2.....

This time we replace n by 2n in your formula

Sum(2n)^2 = 2n(2n + 1)(2n + 2)/6

Testing that out on the 3 terms above 2n(2n + 1)(2n + 2)

{1 to 3}Sum(2n)^2 = 6*7*8/6 = 56

4 + 16 + 36 = 56 so that was OK as well

So we should be able to make a new formula using

S(even) = 1^2 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2.... = S1 - S2

S(even) = [(2n - 1)2n(2n + 1)- 2n(2n + 1)(2n + 2)]/6

S(even) = [2n(2n + 1)/6](2n - 1 - 2n - 2)]/6 = - n(2n + 1)

Test that out with n = 3 applying to the total series having 6 terms

S(even) = - 3(6 + 1) = -21

That is the same result as 35 - 56

You could leave the result that way provided you remember to use

n in your formula when there are 2n terms in your alternating sum.

Or, you could incorporate that into your formula by replacing n by n/2.

Then you get S(even) = - n(n + 1)/2

~~~~~~~~~~~~~~~~~~~~~~~~

But just look at that result again. It should ring a bell.

n(n + 1)/2 is just the sum of n ordinary numbers from 1

It suggests that some easier approach is just waiting to be discovered.

Perhaps we could pair off terms and use the difference of 2 squares

Look at this way of seeing the terms:-

1^2 - 2^2 = -[2^2 - 1^2] = -[(2 - 1)(2 + 1)] = - [1 + 2]

3^2 - 4^2 = -[4^2 - 3^2] = -[(4 - 3)(4 + 3)] = - [3 + 4]

5^2 - 6^2 = -[6^2 - 5^2] = -[(6 - 5)(6 + 5)] = - [5 + 6]

Now we see that the pattern for the sum is

-[1 + 2 + 3 + 4 + 5 + 6] ==> - n(n + 1)/2

Hope you liked that.

P.S. Can you get a similar formula to work for that series with an ODD number of terms ?

Regards - Ian