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

[chwala]

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!

 

[nuuskur]

Consider that it might be redundant to clutter your proof with unnecessary symbols. You write the quantities in different styles: $S_n$ and $f(n)$. Why not $f_n$? The induction is conducted over $\mathbb N$, not the integers. Otherwise, the proof is correct.

Easier to follow: the claim is true for $n=1$. Assume $\sum _{k=1}^n k(3k-1) = n^3+n^2$ for some $n$. Then
$$\begin{align*} \sum _{k=1}^{n+1} k(3k-1) = (n+1)(3n+2) + \sum _{k=1}^n k(3k-1) &= (n+1)(3n+2) + n^3+n^2 \\&= (n+1)(3n+2) + n^2(n+1) \\ &=(n^3+3n^2+3n+1) + (n^2+2n+1) \\ &=(n+1)^3 + (n+1)^2. \end{align*}$$

 

[Mark44]

Consider that it might be redundant to clutter your proof with unnecessary symbols.

Not to mention that each line except the last is an equation that starts with $S_{m+1}$. It's perfectly valid to have one single equation with multiple equal expressions that are chained.

 

[Mark44]

My take;
$\sum_{r=1}^n r(3r-1)=n^3+n^2$

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

The first equation above is what you are to prove. You should explicitly say that.

After you have showed that the proposition is true for the base case -- n = 1 or n = 2, then you should state the induction hypothesis:
Assume that the proposition is true for n = k. I.e.,
$\sum_{r=1}^k r(3r-1)=k^3+k^2$

Now show that the proposition is true for n = k + 1, using the induction hypothesis. I.e., that
$\sum_{r=1}^{k + 1} r(3r-1) = (k+1)^3 + (k+1)^2$.

You can do this like so:
$\sum_{r=1}^{k + 1} r(3r-1) = \dots \text{<intermediate expressions omitted>} = (k+1)^3 + (k+1)^2$

 

[SammyS]

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!

I rather like your approach.

As you stated, showing that $S_1=f(1)$ confirmed that the result was true for $n=1$. (The base step)

You then assumed that $S_m=f(m)$. In this step you are assuming that the result is true for $n=m$ (which should have specified that $m$ was a positive integer, or some equivalent statement). From this assumption, you then showed that $S_{m+1}=f(m+1)$ which shows that the result is true for $n=m+1$. (This is known as the inductive step.)

 

[WWGD]

Since you asked for insights:
$\Sigma r(3r-1) = \Sigma (3r^2 -r) = \Sigma 3r^2 -\Sigma r =3 \Sigma r^2 -\Sigma r$

Both of which have know closed formulars:
$\frac {[3n(n+1)(2n+1)]} {6}$ and $\frac {[n(n+1)}{2}$ respectfully

We can then factor an $n(n+1)/2$ , to get:

$\frac { [3n(n+1)]} {2} [\frac {(2n+1)} {3} -1]] =[ \frac {3n(n+1)}{2} \frac{(2(n-1)}{3}]=n^3 -n^2$

 

[Mark44]

$\Sigma r(3r-1) = \Sigma (3r^2 -r) = \Sigma 3r^2 -\Sigma r =3 \Sigma r^2 -\Sigma r$

Both of which have know closed formulars:
$\frac {[3n(n+1)(2n+1)]} {6}$ and $\frac {[n(n+1)}{2}$ respectfully

Respectfully, that last word should be respectively... (Not to mention formulas/formulae.)

In any case, the problem statement indicated that it should be done using induction.

 

[hutchphd]

In any case, the problem statement indicated that it should be done using induction.

Respectfully, the problem statement said "by induction or otherwise".

I do love inductive proofs however.

 

[WWGD]

Respectfully, that last word should be respectively... (Not to mention formulas/formulae.)

In any case, the problem statement indicated that it should be done using induction.

"Prove by induction or otherwise"
For all intensive purpoises, irregardless. Just joking a bit, Mark.

 

[Mark44]

Respectfully, the problem statement said "by induction or otherwise".

Well, there is that ... The "otherwise" would be a bit easier.

 

[WWGD]

Well, there is that ... The "otherwise" would be a bit easier.

I just realized I could have skipped a step, and could have finished it in two lines.
$\frac {[3n(n+1)(2n+1)]} {6}$ and $\frac {[n(n+1)}{2} = \frac {[n(n+1)(2n+1)]} {2}$ and $\frac {[n(n+1)}{2}$

 

[Mark44]

For all intensive purpoises, irregardless. Just joking a bit, Mark.

Wouldn't that be "for all insensitive porpoises"?
Yep, just having fun.

 

[hutchphd]

The joke I know involves "immortal porpoises" and " transportation across state lions" but I never could deliver a punchline.

 

[renormalize]

The joke I know involves "immortal porpoises" and " transportation across state lions" but I never could deliver a punchline.

staid lions

 

[hutchphd]

(perhaps marginally) different joke! Mine involves a zoo

 

[renormalize]

(perhaps marginally) different joke! Mine involves a zoo

My zoo joke punchline: "Arrested for transporting underage gulls across staid lions for immortal porpoises".

 

[hutchphd]

Almost. Mine was "transporting Mynahs (the birds) across state lions for immortal porpoises" My "state lions" were on loan from the state zoo. Why were your lions "staid"?

 

[WWGD]

(perhaps marginally) different joke! Mine involves a zoo

A zoo? Gesundheit!

 

[renormalize]

Why were your lions "staid"?

Note that this version substitutes "sedated" for "staid" (either word works in the joke):

"A scientist doing research on dolphins discovered that if he fed them a certain type of sea gull, they did not age at all. He kept his secret (and the dolphins) for many years, collecting all the sea gulls he could find while the dolphins continued to live forever, unchanged. One day he went to the beach looking for more sea gulls, but there were none to be found. He had captured them all. So he went to the zoo, where he had heard that they kept a few specimens. He found their cage, snatched several chicks from their nest, and tried to beat a hasty exit from the zoo. On the way out, he snuck through the lion enclosure. Tiptoeing gingerly over a sleeping lion towards the door and freedom, he was immediately arrested and thrown in prison. The charge? Transporting underage gulls across sedate lions for immortal porpoises."

https://arstechnica.com/civis/threads/we-havent-had-a-bad-joke-thread-in-a-while-so.337072/

 

[nuuskur]

Please, this is a homework topic not "Random thoughts part X..".

 

[hutchphd]

I believe we did very well on both counts. Did not wish to offend.