- #1
Nafi Khandaker
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In this problem, Spivak shows how to derive formulas to summations. They start by showing the method for
1^2 + 2^2 + ... + n^2 as follows:
1/k - 1/(k + 1) = 1/(k (k + 1).
Any ideas as to how to derive this result using the method previously given?
1^2 + 2^2 + ... + n^2 as follows:
(k + 1)^3 - k^3 = 3k^2 + 3k + 1
Writing this formula for k = 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.
I understood the method and used it for 1^3 + ... + 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: Writing this formula for k = 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.
1/k - 1/(k + 1) = 1/(k (k + 1).
Any ideas as to how to derive this result using the method previously given?