[SOLVED] proof by induction

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September 9th 2006, 07:14 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
please help me its sooo urgent.

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.
thanks in advance
September 9th 2006, 07:22 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,
$<br /> 1^2+2^2+...+k^2=\frac{k(k+1)(2k+1)}{6}$

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!!!
September 9th 2006, 11:28 PM
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