Algebra 問題

I think you mistype the question, should it be
"Given that 1^2 +2^2 +3^2 + … + n^2 = n(n+1)(2n+1)/6 , find the sum of 1^2+3^2 +5^2 + ...+ 99^2 ."?

If yes, the answer is:
1^2+3^2 +5^2 + ...+ 99^2
= 1^2+2^2+3^2 +4^2 + 5^2 + ...+ 100^2 - (2^2+4^2+...+100^2)
=100*101*201/6 - 2^2(1^2+2^2+...+50^2)
=338350 - 4(50*51*101/6)
=166650

1+3+5+...+99
=(1+2+3+...+50) + (1+2+3+...+49)
= 50(50+1)(100+1)/6 + 49(49+1)(98+1)/6
=83350??

.............咁怪既
係唔係 "1^2 + 2^2 + 3^2 +... +n^2 = n(n+2)(2n+1)/6" 丫??

1+3+5+...+99
= (1^2+2^2+3^2+...+50^2) - (1^2+2^2+3^2+...+49^2)
= 50(50+1)(100+1)/6 + 49(49+1)(98+1)/6
=83350

2007-08-31 01:31:02 補充：
係唔係 "1^2 + 2^2 + 3^2 +... +n^2 = n(n+2)(2n+1)/6" 丫??轉..............................................................+1

2007-08-31 01:33:46 補充：
1+3+5+...+99= (1^2+2^2+3^2+...+50^2) - (1^2+2^2+3^2+...+49^2)= 50(50+1)(100+1)/6 - 49(49+1)(98+1)/6=42925 - 40425=2500

應該係1^2 +2^2 +3^2 + … + n^2 = n(n+1)(2n+1)/6

1^2 +3^2 +5^2 + …..+ 99^2
=1^2 +2^2 +3^2 + … + 100^2 - (2^2+4^2+6^2+...+100^2)
=100(101)(201)/6 - 4 (1^2+2^2+...+50^2)
=67x50x101 - 4(50)(51)(101)/6
=67x50x101 - 4x25x17x101
=1650x101
=166650

2007-08-30 22:23:06 補充：
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