# Series expansions

• Oct 26th 2009, 07:00 AM
orangeiv
Series expansions
I'm attempting this question on binomial distributions:

$\displaystyle 1 + mx + \frac{m(m-1)x^2}{2}$ are the first 3 terms of the series expansion of $\displaystyle (1+x)^m$

Use this to find the 1st 3 terms of the series expansion of $\displaystyle (1+x)^{m+1} (1-2x)^m$

This is what I've done:

1) Substituted (m+1) into the initial expansion to give the expansion for $\displaystyle (1+x)^{m+1}$ :

$\displaystyle 1 + (m+1)x + \frac{m(m+1)x^2}{2}$

2) Used the binomial theorum to expand $\displaystyle (1 - 2x)^m$ :

$\displaystyle 1 - 2mx + {2m(m-1)x^2}$

3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?
• Oct 26th 2009, 07:09 AM
Quote:

Originally Posted by orangeiv
I'm attempting this question on binomial distributions:

$\displaystyle 1 + mx + \frac{m(m-1)x^2}{2}$ are the first 3 terms of the series expansion of $\displaystyle (1+x)^m$

Use this to find the 1st 3 terms of the series expansion of $\displaystyle (1+x)^{m+1} (1-2x)^m$

This is what I've done:

1) Substituted (m+1) into the initial expansion to give the expansion for $\displaystyle (1+x)^{m+1}$ :

$\displaystyle 1 + (m+1)x + \frac{m(m+1)x^2}{2}$

2) Used the binomial theorum to expand $\displaystyle (1 - 2x)^m$ :

$\displaystyle 1 - 2mx + {2m(m-1)x^2}$

3) Is this along the right lines...if so where do I go from here? Or am I doing completely the wrong thing from the start?

HI

I think you are correct . So now multiply the 2 expansions you got .
• Oct 26th 2009, 08:03 AM
orangeiv
Quote:

HI

I think you are correct . So now multiply the 2 expansions you got .

Ok...so we need the $\displaystyle x^0 , x^1 , and x^2$ coefficients...it's horrendous. This is what I got:

$\displaystyle 1 - x(2m-1) + 0.5(m^2 - m)x^2$
• Oct 27th 2009, 06:58 AM
Quote:

Originally Posted by orangeiv
Ok...so we need the $\displaystyle x^0 , x^1 , and x^2$ coefficients...it's horrendous. This is what I got:

$\displaystyle 1 - x(2m-1) + 0.5(m^2 - m)x^2$

I got :

$\displaystyle [1+(m+1)x+\frac{1}{2}m(m+1)x^2+...][1-2mx+2m(m-1)x^2+...]$

$\displaystyle =1-2mx+2m(m-1)x^2+(m+1)x-2m(m+1)x^2+\frac{1}{2}m(m+1)x^2+...$

$\displaystyle =1+(m+1-2m)x+[2m(m-1)-2m(m+1)+\frac{1}{2}m(m+1)x^2+...]$

$\displaystyle =1+(1-m)x+\frac{1}{2}(m^2-7m)x^2+...$