k=1 ending value of k=n-1 eqn= k^3/n^2 How do you deal with the (n-1) ending value when you are writing the sum in closed form?
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Originally Posted by thinair k=1 ending value of k=n-1 eqn= k^3/n^2 How do you deal with the (n-1) ending value when you are writing the sum in closed form? You should know that $\displaystyle \sum_{k=1}^n k^3 = \frac{n^2 (n+1)^2}{4}$. Therefore $\displaystyle \sum_{k=1}^{n-1} k^3 = \frac{(n-1)^2 n^2}{4}$.
Thank you, that is much simpler than what I was trying.
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