Prove: for all n is an element N N = natural numbers n Sigma Notation $\displaystyle (3k^2 - 3k + 1) = n^3 $ k=1
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$\displaystyle *\sum_{k=1}^{n} 3k^2-3k+1 = n^3 $ $\displaystyle \sum_{k=1}^{n+1} 3k^2-3k+1 = \sum_{k=1}^{n} 3k^2-3k+1 + 3(n+1)^2-3(n+1)+ 1 $ $\displaystyle = n^3 + 3(n+1)^2-3(n+1)+ 1 = n^3 + 3n^2 + 3n +1 = (n+1)^3 $
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