What is the value of

`(1^2 +3^2 + ...+171^2) -(2^2 +4^2 +6^2 +...+170^2)?`

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We'll need the following basic facts:

`1^2+2^2+3^2+...+n^2=(n(n+1)(2n+1))/6`

and

`2^2+4^2+...+(2n)^2=(2*1)^2+(2*2)^2+...+(2*n)^2`

`=4*1^2+4*2^2+...+4*n^2`

`=4(1^2+2^2+...+n^2)`

`=(2n(n+1)(2n+1))/3` , using the first formula.

Using these, we can rewrite our problem as

`(1^2+3^2+...+171^2)-(2^2+4^2+...+170^2)`

`=(1^2+2^2+3^2...+171^2)-2(2^2+4^2+6^2+...+170^2)`

`=(171(172)(343))/6-2[(170(86)(171))/3]`

`=14706`

Just remember that in the sum `2^2+4^2+...+170^2,` the last number `170` is equal to `2*85,` so the value we plug in for `n` is `85.`

**Again, the answer is 14706.**

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