Emma Watson
19/03/2017 at 15:47-
→இے๖ۣۜQuỳnh 22/03/2017 at 20:19
We have
Apply (1) to the expression, we have : 13 + 23 + 33 + ... + 20083
= (1 - 1)1(1 + 1) + 1 + (2 - 2)2(2 + 2) + 2 + ....... + (2008 - 2008) 2008 (2008 - 2008) + 2008
= 1 + 2 + 1.2.3 + 3 + 2.3.4 + ........ + 2008 + 2007.2008.2009
= (1 + 2 + 3 + ...... + 2008) + (1.2.3 + 2.3.4 + ...... + 2007.2008.2009)
= 2017036 + 2007.2008.2009.2010 / 4
= The number to mik not count
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¤« 03/04/2018 at 13:40
We have [(n-1)n(n+1)=(n^2-n)(n+1)=(n^2-n)n+n^2-n=n^3-n^2+n^2-n]
[=n^3-n]
[\Rightarrow n^3=(n-1)n(n+1)+n] [(1)]
Apply (1) to the expression, we have : 13 + 23 + 33 + ... + 20083
= (1 - 1)1(1 + 1) + 1 + (2 - 2)2(2 + 2) + 2 + ....... + (2008 - 2008) 2008 (2008 - 2008) + 2008
= 1 + 2 + 1.2.3 + 3 + 2.3.4 + ........ + 2008 + 2007.2008.2009
= (1 + 2 + 3 + ...... + 2008) + (1.2.3 + 2.3.4 + ...... + 2007.2008.2009)
= 2017036 + 2007.2008.2009.2010 / 4
= The number to mik not count
-
Love people Name Jiang 20/03/2017 at 17:43
We have
Apply (1) to the expression, we have : 13 + 23 + 33 + ... + 20083
= (1 - 1)1(1 + 1) + 1 + (2 - 2)2(2 + 2) + 2 + ....... + (2008 - 2008) 2008 (2008 - 2008) + 2008
= 1 + 2 + 1.2.3 + 3 + 2.3.4 + ........ + 2008 + 2007.2008.2009
= (1 + 2 + 3 + ...... + 2008) + (1.2.3 + 2.3.4 + ...... + 2007.2008.2009)
= 2017036 + 2007.2008.2009.2010 / 4
= The number to mik not count