4+8+12+...+4n= 2n(n+1)
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Originally Posted by yingao 4+8+12+...+4n= 2n(n+1) Can you show that it is true for $\displaystyle n=1~?$
4n= 2n(n+1) 4 x1 = 2x1 (1+1) 4 = 4
Originally Posted by yingao 4n= 2n(n+1) 4 x1 = 2x1 (1+1) 4 = 4 O.K. Now if we know that for K>1, 4+8+12+...+4K= 2K(K+1) what can say about 4+8+12+...+4K+4(K+1)?
Last edited by Plato; Apr 1st 2012 at 04:00 PM.
Thanks for your help! I don't know how to go on from here.
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