How Do You Prove the Sum of Sequential Odd Squares Formula?

[charmedbeauty]

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

Homework Statement

show

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

Homework Equations

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). don't I??

 

[tiny-tim]

hi charmedbeauty!

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

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

subtract?

 

[charmedbeauty]

hi charmedbeauty!

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


subtract?

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?

 

[tiny-tim]

but what can I subtract from this

1/3(k+1)(2k+1)(2k+3)+(2k+3)²

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

 

[charmedbeauty]

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

you mean something like this..

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

But I still can't see what to do?

 

[Dick]

you mean something like this..

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

But I still can't see what to do?

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?

 

[charmedbeauty]

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?

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?

 

[Dick]

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?

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?

 

[Ray Vickson]

Homework Statement

show

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

Homework Equations

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). don't I??

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

 

[Whovian]

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

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.

 

[Ray Vickson]

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.

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

RGV