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Given the formula of sequence $$B_{n}=n(n-1)$$ Find the series $S_{n}$ of the sequence, such that $n \geq 0$.

What I did.

$$S_{n} = B_{0} + B_{1} + B_{2} + ... B_{n}$$ $$S_{n} = (0)(0-1) + (1)(1 - 1) + (2)(2 - 1) + ... + n(n-1)$$

Next thing is, I'm stuck. I have no idea how to progress from here. I spend quite some time to reduce this to a closed formula to no avail.

N. F. Taussig
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caramel1995
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    Observe that $$S_n = \sum_{k = 0}^{n} k(k - 1) = \sum_{k = 0}^{n} (k^2 - k) = \sum_{k = 0}^n k^2 - \sum_{k = 0}^{n} k$$ – N. F. Taussig Sep 14 '19 at 16:36

1 Answers1

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Know these two sums:

  • $\displaystyle \sum_{k= 0}^n k = \dfrac{n(n + 1)}2$
  • $\displaystyle \sum_{k = 0}^n k^2 = \dfrac{n(n + 1)(2n + 1)}6$

Then we have: $$\begin{array}{rcl} S_n &=& \displaystyle \sum_{k = 0}^n B_k \\ &=& \displaystyle \sum_{k = 0}^n k(k - 1)\\ &=& \displaystyle \sum_{k = 0}^n (k^2- k)\\ &=& \displaystyle \sum_{k = 0}^n k^2 - \displaystyle \sum_{k = 1}^n k\\ &=& \dfrac{n(n + 1)(2n + 1)}{6} - \dfrac{n(n + 1)}2\\ \end{array}$$

Take out common factors, we have: $$\begin{array}{rcl} S_n &=& \dfrac{n(n + 1)(2n + 1)}{6} - \dfrac{n(n + 1)}2\\ &=& \dfrac{n(n + 1)}{6} \left((2n + 1) - 3\right)\\ &=& \dfrac{n(n + 1)(2n - 2)}{6}\\ &=& \dfrac{n(n - 1)(n + 1)}{3} \end{array}$$

Max Wong
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  • How does the formula of summation $k^2$ came into being??? – caramel1995 Sep 14 '19 at 16:55
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    Notice that $$k^3 - (k - 1)^3 = 3k^2 - 3k + 1$$ The above formula can be proven with binomial theorem. Then we sum both sides from k = 1 to n. We get $$\sum k^3- \sum (k - 1)^3 = 3 \sum k^2 - 3 \sum k + \sum1$$ $$n^3 = 3 \sum k^2 - 3\frac{n(n + 1)}{2} + n $$ Then rearrange and factorise. – Max Wong Sep 14 '19 at 17:06