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Mathematics 19 Online
OpenStudy (anonymous):

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

OpenStudy (tamtoan):

just replace k for n then k+1 for n ..then you get it

OpenStudy (anonymous):

\(S_k:1\times2+2\times3+3\times4+...+k(k+1)=\frac{k(k+1)(k+2)}3\) \(S_{k+1}:1\times2+2\times3+3\times4+...+k(k+1)+(k+1)(k+2)\\ \ \ \ \ \ \ \ \ \ =\frac{k(k+1)(k+2)}3+(k+1)(k+2)=\frac{k(k+1)(k+2)+3(k+1)(k+2)}3\\\ \ \ \ \ \ \ \ \ =\frac{(k+1)(k+2)(k+3)}3\)

OpenStudy (anonymous):

here's another one.. Sn: 1 + 4 + 7 + . . . + (3n - 2) = n(3n - 1)/2

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