Prove that: `1^2-2^2+3^2-4^2+ * * * +n^2=(-1)^(n-1)(n(n+1))/2`  Example 1^2 - 2^2 + .....+11^2=11(11+1)/2=66   How can I prove this mathematically. Thanks

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embizze | High School Teacher | (Level 1) Educator Emeritus

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Lets look at some terms of the sequence:

`1^2=1=(-1)^0(1(2))/2`

`1^2-2^2=-3=(-1)^1(2(3))/2`

`1^2-2^2+3^2=6=(-1)^2(3(4))/2`

`1^2-2^2+3^2-4^2=-10=(-1)^3(4(5))/2`

Thus the sequence is 1,-3,6,-10,15,-21,28,-36,...

We can use mathematical induction:

(1) Base case : n=1 `1^2=1=(1(2))/2` is true.

(2) For the inductive hypothesis:

`1^2-2^2+3^2-4^2+...+n^2=(-1)^(n-1)(n(n+1))/2`

(3) Assume for some `k>=1,kin NN` (k a natural number) that the inductive hypothesis is true.

``(a)If k is odd (so (k+1) is even) we need to show: `1^2-2^2+3^2-4^2+...+k^2-(k+1)^2=-((k+1)(k+2))/2`

We work on the left hand side:

By the inductive hypothesis `1^2-2^2+...+k^2=(k(k+1))/2` ; substituting we get:

`(k(k+1))/2-(k+1)^2`

`=(k^2+k)/2-(k^2+2k+1)`

`=(k^2+k-2k^2-4k-2)/2`   (Get a common denominator and add)

`=(-k^2-3k-2)/2`

`=-((k+1)(k+2))/2` as required.

(b) If k is even, we need to show:

`1^2-2^2+3^2-...-k^2+(k+1)^2=((k+1)(k+2))/2`

Again we work onthe left hand side; by the inductive hypothesis we have:

`1^2-2^2+3^2-...-k^2=-(k(k+1))/2` . Substituting we get:

`-(k(k+1))/2+(k+1)^2`

`=(-k^2-k)/2+k^2+2k+1`

`=(-k^2-k+2k^2+4k+2)/2`

`=(k^2+3k+2)/2`

`=((k+1)(k+2))/2` as required.

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