# Need some help solving this algebra calculation -

1. Nov 2, 2005

### Natasha1

Need some help solving this algebra calculation - URGENT

How can I go from

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

to this

(1/4)n(n+1)(n+2)(n+3)

2. Nov 2, 2005

### CarlB

You've got two polynomials, both 4th order in x. Try putting x=-2, -1, 0, 1, 2 into both of them. Do you get the same number? Then they are equal.

This follows from the calculus of finite differences, but there must be some other reason. If your instructor won't let you reference 19th century mathematics, you might try multiplying the two of them out.

Carl

3. Nov 2, 2005

### Gokul43201

Staff Emeritus
Natasha, it looks like you're trying prove a "sum of fourth powers" theorem by induction...and you are stuck at this final step.

Unless you show what you've done by yourself, people here will not likely respond. And without this, it looks like you're just posting a hw question without any attempt on your part.

So, please post the actual question and the steps you've taken (even if only in a few words) till you got stuck.

PS : Yes, it's me again.

4. Nov 3, 2005

### Natasha1

Right then

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

((12n^3+36n^2+14n+1)) / 12

And I need to get

(1/4)n(n+1)(n+2)(n+3)

Can someone help the simple further steps to take thanks :-)

5. Nov 4, 2005

### Natasha1

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

((4n^3+12n^2+8n+1)) / 4

And I need to get

(1/4)n(n+1)(n+2)(n+3)

Can someone help the simple further steps to take thanks :-)

6. Nov 4, 2005

### HallsofIvy

You can't. The first is a third degree polynomial and the second is a fourth degree polynomial. The can't be equal for all values of n.
In particular, it is easy to see that the first, ((4n^3+12n^2+8n+1)) / 4, is equal to 1/4 when n= 0 while the second,(1/4)n(n+1)(n+2)(n+3), is equal
to 0.

7. Nov 4, 2005

### Natasha1

Could someone just help me please, can someone spot where I am going wrong please?

((n^2+(n+1)^2)/4) + 3 [(1/6)n(n+1)(2n+1)] + 2 [(1/2)n(n+1)]

After putting all the terms to a common denominator. Then expanding out, and simplifying the above I get

((3n^2+3(n+1)^2)/12) + 6 [(n(n+1)(2n+1))/12] + 12 [(n(n+1))/12]
then
(3n^2+3(n^2+2n+1) + 6n(2n^2+n+2n+1)+12n(n+1))/12
then
(3n^2+3n^2+6n+3 + 12n^3+6n^2+12n^2+6n+12n^2+12n)/12
then
(12n^3+36n^2+24n+3)/12
then
((4n^3+12n^2+8n+1)) / 4