BrainDen.com - Brain Teasers

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• 0 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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• 0 P.S. bushindo, perfectly right.

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• 0

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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