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Given f(n)=1/(n+1)+1/(n+2)+...+1/(3n+1). When you want to prove the question using the mathematical induction method. For step 2, if n=k, you want to validate when n=k+1, then the relationship between f(k) and f(k+1) will be?

Thank you so much in advance!!

yasbib555  Jul 15, 2018
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I do not really understand your question, I think it is incomplete but perhaps you only want this part of it.

Given f(n)=1/(n+1)+1/(n+2)+...+1/(3n+1). When you want to prove the question using the mathematical induction method. For step 2, if n=k, you want to validate when n=k+1, then the relationship between f(k) and f(k+1) will be?

$$f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(n)=\frac{1}{n+1}+\frac{1}{n+2}+....+\frac{1}{2n}+ ....+\frac{1}{3n}+\frac{1}{3n+1}\\ f(k)=\frac{1}{k+1}+\frac{1}{k+2}+....+\frac{1}{2k}+ ....+\frac{1}{3k}+\frac{1}{3k+1}\\~\\ f(k+1)=\frac{1}{k+2}+.......\frac{1}{2k+2}+......\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\frac{1}{k+2}+.....+\frac{1}{3k+1}+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[\frac{1}{k+2}+.....+\frac{1}{3k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=\left[f(k)-\frac{1}{k+1}\right]+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}\\ f(k+1)=f(k)+\frac{1}{3k+2}+\frac{1}{3k+3}+\frac{1}{3k+4}-\frac{1}{k+1}\\$$

check:

$$f(1)=\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\\ f(2)=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\\ f(2)=f(1)+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}-\frac{1}{2}$$

Well that looks promising.  :)

Melody  Jul 15, 2018
edited by Melody  Jul 15, 2018