Find the value of ' 1^2-2^2+3^2-4^2+5^2-6^2+.................-98^2+99^2 '

Asked by  | 15th Dec, 2008, 09:48: PM

Expert Answer:

12-22+32-42+.....-982+992

=(12+32+...+992) -(22+42+...+982)

=(12+22+32+42+...+982+992) - 2 (22+42+...+982)

= (sum of squares of first 99 natural numbers) -2 . 22 (sum of squares of first 49 natural numbers)

= [99. (99+1) . (2.99+1) /6] - 8. [49. (49+1).(2.49+1)/6]

Answered by  | 17th Dec, 2008, 11:48: AM

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