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Find a simple closed form expression for

the alternating series of squares,

n2-(n-1)2+(n-2)2-(n-3)2+...+42-32+22-12.

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Posted · Report post

Find a simple closed form expression for

the alternating series of squares,

n2-(n-1)2+(n-2)2-(n-3)2+...+42-32+22-12.

It should be equal to

n2/2 + n/2

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Posted · Report post

It should be equal to

n2/2 + n/2

think you forgot the alternater {(n2+n)/2}(-1)n

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Posted · Report post

1. The question implies that n is even. Had n benn odd, the last term "1" would have been "+1" instead of "-1"

2. Since n is even, it is possible to pair all the terms like

[n^2 - (n-1)^2] + [(n-2)^2 - (n-3)^2] + ...

3. This is = [2n - 1] + [2n - 5] + ... +[2n - (2n - 3)] i.e. n/2 terms

4. = 2n * n/2 - [1 + 5 + 9 + ... extending to n/2 terms

5. = n^2 - n(n - 1)/2

6. = n(n + 1)/2

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Posted · Report post

P.S. bushindo, perfectly right.

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Posted · Report post

DOH! caught me sp - nice one. and thanks for calling me out Mukul Verma. should have known better (not to get the answer but to have doubted bushindo).

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Posted (edited) · Report post

the difference b/w n^2 and (n-1)^2 is 2n-1, so the expr. =

(2n-1)+(2n-5)+........+3 which is an AP, so

sum = n/2(n+1)

Edited by dark_magician_92
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