# [SOLVED] proof by induction

• Sep 9th 2006, 08:14 PM
princess
[SOLVED] proof by induction
proof by induction; you follow the three steps ive got that much but i dont know how to do it when the numbers are squared.
for example...
1squared + 2squared + 3 squared + ........... + nsquared = 1/6 (n+1)(2n+1)

i can complete the first step (prove its true for 1)
1/6 (1+1)(2+1) = 1/6 x 2 x 3
= 1
1squared = 1

step 2 is to assume that it is true for n=k i think i am getting lost on this step.
i have reached
1/6k (k+1)(2k+1) = m (i think that part is right)

however step three is to prove it is true for n=k+1
this is completely loosing me and it is due tomorrow im totally lost

p.s i dont want the answer just some help if that is possible, im more interested in how the answer is recieved then the actual answer that it gives.
• Sep 9th 2006, 08:22 PM
ThePerfectHacker
The equation editor (LaTeX) is curretly disabled because the site is under repair. Please try to understand this mess.
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The first part you got correct, you checked $n=1$.

The second part is,
$
1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

Now, to get the next case you add $(k+1)^2$ to both sides, thus,

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

Now, add the fraction on the right hand side,
$1^2+2^2+...+k^2+(k+1)^2=\frac{k(k+1)(2k+1)+6(k+1)^ 2}{6}$
Thus, the right hand side, can be factored
$\frac{(k+1)[k(2k+1)+6(k+1)]{6}$
Thus,
$\frac{(k+1)[2k^2+k+6k+6]{6}$
Thus,
$\frac{(k+1)(k+2)(2k+3)}{6}$
It works!!!
• Sep 10th 2006, 12:28 AM
CaptainBlack
Quote:

Originally Posted by princess
proof by induction; you follow the three steps ive got that much but i dont know how to do it when the numbers are squared.
for example...
1squared + 2squared + 3 squared + ........... + nsquared = 1/6 (n+1)(2n+1)

i can complete the first step (prove its true for 1)
1/6 (1+1)(2+1) = 1/6 x 2 x 3
= 1
1squared = 1

step 2 is to assume that it is true for n=k i think i am getting lost on this step.
i have reached
1/6k (k+1)(2k+1) = m (i think that part is right)

If you assume the result true for n=k, you have:

1^2 + 2^2 + ... + k^2 = (k+1)(2k+1)/6,

from this you need to show that:

1^2 + 2^2 + ... + k^2 + (k+1)^2 = (k+2)(2k+5)/6,

which is what the result would give if n=k+1. You would usually proceed
by noting by the assumption that:

1^2 + 2^2 + ... + k^2 + (k+1)^2 = (k+1)(2k+1)/6 + (k+1)^2.

Then you rearrange the right hand side of this last equation until
it looks like the right hand side of the previous equation, and then
this stage is done.

RonL