In this problem, Spivak shows how to derive formulas to summations. They start by showing the method for(adsbygoogle = window.adsbygoogle || []).push({});

1^2 + 2^2 + ... +n^2 as follows:

(I understood the method and used it for 1^3 + ... +k+ 1)^3 -k^3 = 3k^2 + 3k+ 1

Writing this formula fork= 1, 2, ..., n and adding, we obtain

2^3 - 1^3 = 3*1^2 + 3*1 + 1

3^3 - 2^3 = 3*2^2 + 3*2 + 1

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(n+ 1)^3 -n^3 = 3*n^2 + 3*n+ 1

(n+ 1)^3 - 1 = 3[1^2 + ... +n^2] + 3[1 + ... +n] +n.

n^3 and 1^4 + ... +n^4, but I am completely lost as to how to apply this to 1/(1*2) + 1/(2*3) + ... 1/(n(n+ 1)). The book's answer doesn't help much:

1/k- 1/(k+ 1) = 1/(k(k+ 1).

Any ideas as to how to derive this result using the method previously given?

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# Calculus by Spivak, Chapter 2, Problem 6, Part 3

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