We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
273
1
avatar+283 

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!!

 Jul 15, 2018
 #1
avatar+102761 
+1

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.  :)

 Jul 15, 2018
edited by Melody  Jul 15, 2018

14 Online Users