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- Homework Statement
- Prove by induction or otherwise, that ##\sum_{r=1}^n r(3r-1)=n^3+n^2##

- Relevant Equations
- understanding of induction

My take;

##\sum_{r=1}^n r(3r-1)=n^3+n^2##

Let ##S_n=\sum_{r=1}^n r(3r-1)## and ##f(n)=n^3+n^2##

then, ##S_1 = 2## and ##f(1)=2##

hence the result is true when ##n=1##

Assume that ## S_m = f(m)## for some integer ##m## then,

## S_{m+1}= (m+1)(3(m+1)-1) +S_m ##

## S_{m+1}= (m+1)(3(m+1)-1) +m^3+m^2 ##

## S_{m+1}=m^3+m^2 +(m+1)(3m+2)##

## S_{m+1}= m^3+m^2+3m^2+5m+2 ##

## S_{m+1}=m^3+3m^2+3m+1+(m^2+2m+1)##

## S_{m+1}=(m+1)^3+(m+1)^2##

## S_{m+1}=f(m+1)##

##S_1=f(1)## , ##S_2=f(2)##

implying that, ##S_3=f(3)## for any integer ##n## thus, the result is true for all ##n##.

Your insight or any other approach welcome! Cheers!

##\sum_{r=1}^n r(3r-1)=n^3+n^2##

Let ##S_n=\sum_{r=1}^n r(3r-1)## and ##f(n)=n^3+n^2##

then, ##S_1 = 2## and ##f(1)=2##

hence the result is true when ##n=1##

Assume that ## S_m = f(m)## for some integer ##m## then,

## S_{m+1}= (m+1)(3(m+1)-1) +S_m ##

## S_{m+1}= (m+1)(3(m+1)-1) +m^3+m^2 ##

## S_{m+1}=m^3+m^2 +(m+1)(3m+2)##

## S_{m+1}= m^3+m^2+3m^2+5m+2 ##

## S_{m+1}=m^3+3m^2+3m+1+(m^2+2m+1)##

## S_{m+1}=(m+1)^3+(m+1)^2##

## S_{m+1}=f(m+1)##

##S_1=f(1)## , ##S_2=f(2)##

implying that, ##S_3=f(3)## for any integer ##n## thus, the result is true for all ##n##.

Your insight or any other approach welcome! Cheers!

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