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# help sum

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Find the closed form sum of 1*2*3 + 2*3*4 + 3*4*5 + ... + n(n + 1)(n + 2).

Apr 16, 2022

#1
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$$\begin{array}{rcl} \displaystyle\sum_{k = 1}^n k(k + 1)(k + 2) &=& \displaystyle\sum_{k = 1}^n (k^3 + 3k^2 + 2k)\\ &=& \displaystyle\sum_{k = 1}^n k^3 + 3\displaystyle\sum_{k = 1}^nk^2 + 2\displaystyle\sum_{k = 1}^nk\\ &=& \left(\dfrac{n(n + 1)}2\right)^2 + 3\left(\dfrac{n(n + 1)(2n + 1)}6\right) + 2\left(\dfrac{n(n + 1)}2\right)\\ &=& \dfrac{n(n + 1)\left(n(n + 1) + 2(2n + 1)+4\right)}{4}\\ &=& \dfrac{n(n+ 1)(n^2 + 5n + 6)}{4}\\ &=& \dfrac{n(n + 1)(n + 2)(n + 3)}4 \end{array}$$

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Apr 17, 2022

$$\begin{array}{rcl} \displaystyle\sum_{k = 1}^n k(k + 1)(k + 2) &=& \displaystyle\sum_{k = 1}^n (k^3 + 3k^2 + 2k)\\ &=& \displaystyle\sum_{k = 1}^n k^3 + 3\displaystyle\sum_{k = 1}^nk^2 + 2\displaystyle\sum_{k = 1}^nk\\ &=& \left(\dfrac{n(n + 1)}2\right)^2 + 3\left(\dfrac{n(n + 1)(2n + 1)}6\right) + 2\left(\dfrac{n(n + 1)}2\right)\\ &=& \dfrac{n(n + 1)\left(n(n + 1) + 2(2n + 1)+4\right)}{4}\\ &=& \dfrac{n(n+ 1)(n^2 + 5n + 6)}{4}\\ &=& \dfrac{n(n + 1)(n + 2)(n + 3)}4 \end{array}$$