# Show that 1^2+3^2+ (2n+1)^2

1. Apr 27, 2012

### charmedbeauty

Show that 1^2+3^2+....(2n+1)^2

1. The problem statement, all variables and given/known data

show

1^2+3^2+....+(2n+1)^2=1/3(n+1)(2n+1)(2n+3) whenever n is an element of natural numbers.

2. Relevant equations

3. The attempt at a solution

So I started with the basis step of proving true for n=1

so 1^2+.....+(2(1)+1)^2=1/3(1+1)(2+1)(2+3)

so 10 = 1/3(2)(3)(5) =10 so true for n=1

Assume true for n=k

Show it true for n=k+1

so 1^2+3^2+....(2k+1)^2+(2k+3)= 1/3(n+1)(2n+1)(2n+3)+(2k+3)^2

as 1/3(n+1)(2n+1)(2n+3)=1^2+3^2+....(2k+1)^2

But I don't know where to go from here??

I have to get it in the form

1/3(odd integer)(odd integer)(odd integer). dont I??

2. Apr 27, 2012

### tiny-tim

hi charmedbeauty!

(try using the X2 button just above the Reply box )
subtract?

3. Apr 27, 2012

### charmedbeauty

Re: Show that 1^2+3^2+....(2n+1)^2

but what can I subtract from this

1/3(k+1)(2k+1)(2k+3)+(2k+3)2

if I expand out I get

1/3(4k3+12k2+11k+3)+1/3(12k2+36k+27)

rounding up terms..

1/3(4k3+24k2+47k+30)

where does the subtraction come from? I thought the two original terms above were separated by addition?

4. Apr 27, 2012

### tiny-tim

it'll help to factor out (2k+3) first

5. Apr 27, 2012

### charmedbeauty

Re: Show that 1^2+3^2+....(2n+1)^2

you mean something like this..

(2k+3)(1/3(k+1)(2k+1)+(2k+3))

But I still cant see what to do?

6. Apr 27, 2012

### Dick

Re: Show that 1^2+3^2+....(2n+1)^2

You want to show that result is the same as putting n to be n+1 in (1/3)(n+1)(2n+1)(2n+3). That's (1/3)(n+2)(2n+3)(2n+5). Is it?

7. Apr 27, 2012

### charmedbeauty

Re: Show that 1^2+3^2+....(2n+1)^2

Yes I can see but the problem I keep on having is I'm always left with that stupid addition sign, how do I make it go away?

8. Apr 27, 2012

### Dick

Re: Show that 1^2+3^2+....(2n+1)^2

If you expand out (1/3)(n+2)(2n+3)(2n+5) do you get the same thing as (1/3)(n+1)(2n+1)(2n+3)+(2n+3)^2?

9. Apr 27, 2012

### Ray Vickson

Re: Show that 1^2+3^2+....(2n+1)^2

Your problem notation did not make clear whether you were dealing with problem (1) or (2) below:
(1) sum_{j <= 2n+1} j^2 (all numbers from 1 to 2n+1), or
(2) sum_{j <= 2n+1, j ODD} j^2 (odd numbers only from 1 to 2n+1).

RGV

10. Apr 27, 2012

### Whovian

Re: Show that 1^2+3^2+....(2n+1)^2

Yes it did. The first two terms were 1^2+3^2, and that's enough to imply the second, as the first's second term would be 2^2, not 3^2.

11. Apr 27, 2012

### Ray Vickson

Re: Show that 1^2+3^2+....(2n+1)^2

You'r right. Sorry. I need my morning coffee.

RGV