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A statement Sn about the positive integers is given. Write statements Sk and Sk+1, simplifying Sk+1 completely. Show your work. Sn: 1 ∙ 2 + 2 ∙ 3 + 3 ∙ 4 + . . . + n(n + 1) = [n(n + 1)(n + 2)]/3

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coolstick
Alright, for the sum part, it really seems like it's x*(x+1) +(x+1)*(x+2)repeating forever. Therefore, it is  ∑ (1 at the bottom, n at the top) k*(k+1)=[n(n + 1)(n + 2)]/3

Answer with explanation:

[tex]S_{n}= 1 \times 2 + 2 \times 3 + 3 \times 4 + . . . + n\times(n + 1) = \frac{n(n + 1)(n + 2)}{3}\\\\S_{k}= 1 \times 2 + 2 \times 3 + 3 \times 4 + . . . + k\times(k + 1) = \frac{k \times(k + 1)(k + 2)}{3}\\\\S_{k+1}= 1 \times 2 + 2 \times 3 + 3 \times 4 + . . . + k\times(k + 1)+(k+1) \times (k+2)=\frac{(k+1) \times(k +1+ 1)(k +1+ 2)}{3}\\\\=\frac{(k+1) \times(k +2)\times(k +3)}{3}[/tex]