September 19th, 2016, 10:02 AM  #1 
Member Joined: May 2016 From: Ireland Posts: 92 Thanks: 1  induction
prove by induction for all positive integers n: 1+5+9+13+........+(4n3)= n/2(4n2) i tried this by trying to prove n/2(4n2)+ (4(k+1)3) = k+1/2(4(k+1)2) but it did not work out for me. 
September 19th, 2016, 10:16 AM  #2 
Senior Member Joined: Dec 2012 From: Hong Kong Posts: 789 Thanks: 285 Math Focus: Stochastic processes, statistical inference, data mining, computational linguistics 
You mean the second step of the induction? Assume $\displaystyle 1 + 5 + 9 + ... + (4k3) = \frac{k(4k2)}{2}$. Then $\displaystyle 1 + 5 + 9 + ... + (4k3) + [4(k+1)3] = \frac{k(4k2)}{2} + 4(k+1)3 = \frac{k(4k2) + 8(k+1)  6}{2} = \frac{4k^22k + 8k+8  6}{2} = \frac{4k^2+6k+2}{2}=\frac{(k+1)(4k+2)}{2}=\frac{(k +1)(4(k+1)2)}{2}$. 
September 19th, 2016, 10:51 AM  #3 
Senior Member Joined: Aug 2016 From: morocco Posts: 273 Thanks: 32 
For the second step, you must say: let k be an integer >=1 such that 1+5+...(4k3)=.... (IH) let's prove that 1+5+....(4k3)+(4(k+1)3)=...... then we use the induction hypothesis. 

Tags 
induction 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Induction 3  med1student  Algebra  10  September 4th, 2015 08:09 PM 
Induction  Tommy_Gun  Algebra  14  June 1st, 2012 11:52 PM 
Induction 2  TheTree  Algebra  1  May 28th, 2012 10:54 AM 
induction  gaussrelatz  Algebra  4  September 28th, 2011 08:55 PM 
Induction  Tommy_Gun  Number Theory  0  January 1st, 1970 12:00 AM 