1. Induction

I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

I'm following the example here

The bit that throws me is how they got from:
$= (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)$
To:
$= \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]$

Doesn't make sense to me how they got rid of the + and the power of 2.

Thanks.

2. Originally Posted by alexgeek
I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

I'm following the example here

The bit that throws me is how they got from:
$= (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)$
To:
$= \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]$

Doesn't make sense to me how they got rid of the + and the power of 2.

Thanks.
hi
they factorized

3. Originally Posted by alexgeek
I'm trying to teach myself induction, I get the idea behind it but theres always some weird jump in algebra at some point that throws me in the examples.

I'm following the example here

The bit that throws me is how they got from:
$= (n+1)^2 + \frac{1}{6}n(n+1)(2n+1)$
To:
$= \frac{1}{6}(n+1)[6(n+1) + n(2n+1)]$

Doesn't make sense to me how they got rid of the + and the power of 2.

Thanks.
Man, if you got the rest of what they said, and this is the only thing that you have trouble with then consider yourself lucky. The step that "doesn't make sense" to you is simple factoring. Can you see?

4. Oh right ha, just found it hard to spot.
I'm gunna try and see if I can do one of the examples.
Thanks

5. I'm trying to prove the sum of an arithmetic series,
$P(n): \sum_{i=1}^{n} a+(n-1)d = \frac{n}{2}(2a+(n-1)d)$
for $a,d,n > 0$

I've proved it for
$Let a = d = 1$
$P(1) = \frac{1}{2}(2 + 0) = \frac{2}{2} = 1$

I'm getting stuck on proving
$\sum_{i=1}^{n+1} = a+(n-1)d + \sum_{i=1}^{n}$
$\frac{n+1}{2}(2a + ( (n+1) - 1)d ) = a + (n-1)d + \frac{n}{2}(2a + (n-1)d)$

Closest I've got is:
$\frac{2an + 2a}{2} + \frac{n+1}{2}[((n+1)-1)d] = a + (n-1)d + \frac{2an}{2} + \frac{n}{2} ((n-1)d)$

Which gets rid of an and a on both sides, but everything else I do seems to fail leaving me with unequal amounts of d's and n's on each side.

Anyone got any ideas? thanks

6. do u mean $\sum_{i=1}^{n}a_i$ or $\sum_{i=1}^{n}i$.

7. Hi alex,

why not do it this way ?

$\sum_{i=1}^n\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)$

$\sum_{i=1}^{n+1}\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)+a+nd$

which should equal

$\frac{n+1}{2}\left(2a+nd\right)$

Proof

$\frac{n}{2}\left(2a+[n-1]d\right)+a+nd=\frac{n}{2}(2a+nd)+\frac{n}{2}(-d)+a+nd$

$=\frac{n}{2}(2a+nd)-\frac{nd}{2}+\frac{2a}{2}+\frac{2nd}{2}$

$=\frac{n}{2}(2a+nd)+\frac{nd}{2}+\frac{2a}{2}$

$=\frac{n}{2}(2a+nd)+\frac{1}{2}(2a+nd)=\frac{n+1}{ 2}(2a+nd)$

8. Edit: nevermind, this post is not appropriate.

9. Originally Posted by Archie Meade
Hi alex,

why not do it this way ?

$\sum_{i=1}^n\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)$

$\sum_{i=1}^{n+1}\left(a+[i-1]d\right)=\frac{n}{2}\left(2a+[n-1]d\right)+a+nd$

which should equal

$\frac{n+1}{2}\left(2a+nd\right)$

Proof

$\frac{n}{2}\left(2a+[n-1]d\right)+a+nd=\frac{n}{2}(2a+nd)+\frac{n}{2}(-d)+a+nd$

$=\frac{n}{2}(2a+nd)-\frac{nd}{2}+\frac{2a}{2}+\frac{2nd}{2}$

$=\frac{n}{2}(2a+nd)+\frac{nd}{2}+\frac{2a}{2}$

$=\frac{n}{2}(2a+nd)+\frac{1}{2}(2a+nd)=\frac{n+1}{ 2}(2a+nd)$

ahh, so when I was adding the extra term to the right, I forgot to add 1 to n.
Thanks loads for all that, couldn't find it on the net.
I'll see if I can do geometric series now