# Thread: Please help me solve this Algebaric Problem

1. ## Please help me solve this Algebaric Problem

I want to open the brackets of this problem

-(1/2)((x/2)+(x^2/4)+(x^3/12)+.................)^2+(1/3)((x/2)+(x^2/4)+..............)^3

The answer to this should be -(x^2/8)-(x^3/8)-(x^4/32)-(x^4/24)-........+(x^3/24)+(x^4/16)+..........

This is a part of a expansion problem whose original question is log(1+e^x)
I know we have to use the binomial theorem but dont know how.
Because in (a+b)^n there are only two terms but in the first bracket there are three terms.
Please help me with this as i have my exams on this within few days. Thank You.

2. Originally Posted by sajoson
I want to open the brackets of this problem

-(1/2)((x/2)+(x^2/4)+(x^3/12)+.................)^2+(1/3)((x/2)+(x^2/4)+..............)^3

The answer to this should be -(x^2/8)-(x^3/8)-(x^4/32)-(x^4/24)-........+(x^3/24)+(x^4/16)+..........

This is a part of a expansion problem whose original question is log(1+e^x)
I know we have to use the binomial theorem but dont know how.
Because in (a+b)^n there are only two terms but in the first bracket there are three terms.
Please help me with this as i have my exams on this within few days. Thank You.
You might find this helpful:
$\displaystyle (a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2ac + 2bc$
as you can verify by multiplying out the expression. Can you spot the pattern for squaring a multinomial?

-Dan

3. Hey thanks Dan,

Can you give me the expression for (a+b+c+d)^2, (a+b)^3 & (a+b+c)^3 too.

Thanks.

4. Originally Posted by sajoson
Hey thanks dan,

Can you give me the expression for (a+b+c+d)^2, (a+b)^3 & (a+b+c)^3 too.
(Shaking my finger) Tsk, tsk! You should be able to generate these by yourself! They aren't hard to do, they just take a little persistence.

$\displaystyle (a + b + c + d)^2 = a^2 + b^2 + c^2 + d^2 + 2ab + 2ac + 2ad + 2bc + 2bd + 2cd$

$\displaystyle (a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

$\displaystyle (a + b + c)^3$
= $\displaystyle a^3 + b^3 + c^3$
+ $\displaystyle 3a^2b + 3a^2c$
+ $\displaystyle 3ab^2 + 3ac^2$
+ $\displaystyle 3b^2c$
+ $\displaystyle 3bc^2$
+ $\displaystyle 6abc$
(Hopefully breaking this up like this will help you see the pattern.)

The smaller products aren't hard to work out or spot patterns in to help you remember them. But these are merely specific examples of a more general expansion, called the "multinomial theorem."

Hope it helps!

-Dan

5. Originally Posted by sajoson
Hey thanks Dan,

Can you give me the expression for (a+b+c+d)^2, (a+b)^3 & (a+b+c)^3 too.

Thanks.
Hello,

you only have to multiply two (or three) bracketed terms. I assume that you know how to do this. Thus i give you only the final result:

$\displaystyle (a+b+c+d)^2=a^2 + 2·a·b + 2·a·c + 2·a·d + b^2 + 2·b·c + 2·b·d + c^2 + 2·c·d + d^2$

$\displaystyle (a+b)^3=a^3+3a^2b+3ab^2+b^3$

$\displaystyle (a+b+c)^3=a^3 + 3·a^2·b + 3·a^2·c + 3·a·b^2 + 6·a·b·c + 3·a·c^2 + b^3 + 3·b^2·c + 3·b·c^2 + c^3$

EB

6. I only needed the final result. thanks. Thank you very much Dan. You have been really helpful.