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Sorry about bad englsih Guys, i have this:

n n-1 ( Σ 3k²-k) + ( Σ 2k-3k²) k=1 k=0

So, the limits of the first one are: k=1 and n and of the second one are k=0 and n-1. To proced, i have to adjust the limits until both have the same limits (k=0 and n), and thats my problem, i dont know exactaly how to manipulate this limits. It would be correct do something like this?

n n ( Σ 3k²-k) -> ( Σ 3k²-k+k) k=1 k=0

In this example i subtract the value of k in the limit and add this value into the summation. Would apreciate your help!

GGirotto
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  • when replacing $k=1$ with $k=0$ in the limits, replace $k$ with $k+1$ in the summation. Notice both give the same starting values and subsequent values. $\sum_{k=1}^n 3k^2-k = \sum_{k=0}^n 3(k+1)^2-(k+1)$ – MAM Apr 25 '15 at 21:36
  • So, if i change the k valeu from 3 to 0, it would add k+3 in the summation? and for the n, how dows it works? – GGirotto Apr 25 '15 at 21:39

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Hint:

Well all you have to do is manipulate the second one:

$$\sum _{ k=0 }^{ n-1 }{ (2k-{ 3k }^{ 2 }) } =\sum _{ k=1 }^{ n }{ (3k^2-k)} = \sum _{ k=0 }^{ n }{(3(k+1)^2−(k+1))}$$