# Thread: Quick mathematical induction question

1. ## Quick mathematical induction question

The problem

n
Σi^2 = n(n + 1)(2n + 1)/6
i=1

The base case is correct. Gona get to the meat of the problem

P(k) -------> P(k +1)
1^2 + 2^2 + 3^2 ......+k^2 = k(k +1)(2k+ 1)/6

Now showing k +1
1^2 + 2^2 + 3^2 ......+k^2 + (K +1)^2 = k+1((k +1)+1)((2k+ 1)+1)/6

RHS = (2k^3 + 7k^2 + 10k + 6)/6

LHS =
1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2

Before I go on was I right in mulitplyong everything out with k +1 on the RHS so i could math it up with the LHS? in bold

Thank you!!

2. It's bit unclear what you've mentioned

what have you done before the last line?

3. Originally Posted by BAdhi
It's bit unclear what you've mentioned

what have you done before the last line?
Thats the right hand sides multiplied together when I replaced the k with k +1

4. Originally Posted by Thetheorycase
LHS =
1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2
You have left out an important equals sign! You mean, of course,
$\displaystyle 1^2+ 2^2+ 3^2+ ...+ k^2+ (k+1)^2= k(k+1)(2k+1)/6+ (k+1)^2$

Imstead of multiplying that out, factoring k+ 1 out of the right side you get $\displaystyle (k+1)(k(2k+1)/6+ k+1)$
$\displaystyle k(2k+1)/6+ k+1= [k(2k+1)+ 6(k+1)]/6= [2k^2+ 7k+ 6]/6$
can you factor that numerator?

5. Originally Posted by Thetheorycase
The problem

n
Σi^2 = n(n + 1)(2n + 1)/6
i=1

The base case is correct. Gona get to the meat of the problem

P(k) -------> P(k +1)
1^2 + 2^2 + 3^2 ......+k^2 = k(k +1)(2k+ 1)/6

Now showing k +1
1^2 + 2^2 + 3^2 ......+k^2 + (K +1)^2 = k+1((k +1)+1)((2k+ 1)+1)/6

RHS = (2k^3 + 7k^2 + 10k + 6)/6

LHS =
1^2 + 2^2 + 3^2 ......+k(k +1)(2k+ 1)/6 + (K +1)^2

Before I go on was I right in mulitplyong everything out with k +1 on the RHS so i could math it up with the LHS? in bold

Thank you!!

P(k+1)

It looks like you multiplied out

$\displaystyle \displaystyle\frac{\left(k^2+2k+2\right)\left(2k+3 )}{6}=\frac{2k^3+7k^2+10k+6}{6}$

which means you haven't correctly multiplied out the RHS, which should have read

$\displaystyle \displaystyle\frac{(k+1)[(k+1)+1][2(k+1)+1]}{6}=\frac{(k+1)(k+2)(2k+3)}{6}=\frac{\left(k^2+3k +2\right)(2k+3)}{6}$

You don't have to multiply the factors out

You can keep RHS=$\displaystyle \displaystyle\frac{(k+1)(k+2)(2k+3)}{6}$

Instead you can show the LHS is this

$\displaystyle \displaystyle\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{ k(k+1)(2k+1)}{6}+\frac{6(k+1)^2}{6}$

Both terms have a common factor of $\displaystyle \displaystyle\frac{k+1}{6}$

$\displaystyle \displaystyle\frac{k+1}{6}\;[{k(2k+1)+6(k+1)]=\frac{k+1}{6}\left(2k^2+7k+6\right)$

and you can finish up by factoring $\displaystyle 2k^2+7k+6$