Question: Find Pk+1 for the given Pk. Pk = k/2 (3k-1) So basically for every k, I put k+1. Therefore: (k+1)/(2) [3(k+1)-1] This may sound stupid, but how do I simplify from here? Any help is appreciated.

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Originally Posted by iluvmathbutitshard Question: Find Pk+1 for the given Pk. Pk = k/2 (3k-1) So basically for every k, I put k+1. Therefore: (k+1)/(2) [3(k+1)-1] This may sound stupid, but how do I simplify from here? Any help is appreciated. $\displaystyle \frac{k+1}{2} \left[3(k+1) - 1\right] $ $\displaystyle \frac{k+1}{2} \left(3k+3-1\right)$ $\displaystyle \frac{k+1}{2} \left(3k+2\right)$ $\displaystyle \frac{(k+1)(3k+2)}{2} $