# Triple term binomial expansion

• Dec 14th 2010, 08:37 PM
arccos
Triple term binomial expansion
In the expansion $(1 + ax + bx^2)^5$ the coefficients of $x$ and $x^2$ are 10 and 25 respectively, where a and b are constants.

Find the values of a and b.

I have tried to rewrite the equation in the form $((1 + ax) + bx^2)^5$ and prepare it for binomial expansion but I am getting something like
$\displaystyle 5 \choose 0$ $\displaystyle (1+ax)^5(bx^2)^0 +$ $\displaystyle 5 \choose 1$ $\displaystyle (1+ax)^4(bx^2)^1 + ......$

Which we can use to solve for a and b but it is extremely long and tedious and I am sure there must be an easier way to get around it. Any help at all is appreciated.
• Dec 14th 2010, 11:09 PM
FernandoRevilla
Use:

$(u+v+w)^5=\displaystyle\sum_{\alpha +\beta +\gamma=5}\dfrac{5!}{\alpha!\beta!\gamma!}u^{\alph a}v^{\beta}w^{\gamma}$

$\alpha,\beta,\gamma$ non-negative integers.

Fernando Revilla
• Dec 15th 2010, 12:30 AM
arccos
the trinomial theorem is not in my syllabus. is there another way around it?
• Dec 15th 2010, 01:37 AM
FernandoRevilla
Quote:

Originally Posted by arccos
the trinomial theorem is not in my syllabus. is there another way around it?

In that case, use the tedious method. :)

Fernando Revilla
• Dec 15th 2010, 01:39 AM
TheCoffeeMachine
$\displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+...$.

By inspection, the coefficient of $x$ is $5a$ and that of $x^2$ is $10a^2+5b$, thus we have:

$5a = 10 \Rightarrow a = 2$ and $10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3$.
• Dec 15th 2010, 03:37 AM
CaptainBlack
Quote:

Originally Posted by TheCoffeeMachine
$\displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+...$.

By inspection, the coefficient of $x$ is $5a$ and that of $x^2$ is $10a^2+5b$, thus we have:

$5a = 10 \Rightarrow a = 2$ and $10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3$.

You should point out that the ellipsis ... contains only terms of order greater than 2.

CB
• Dec 15th 2010, 03:44 AM
TheCoffeeMachine
Quote:

Originally Posted by CaptainBlack
You should point out that the ellipsis ... contains only terms of order greater than 2.

I should have, indeed. Thanks.
• Dec 15th 2010, 05:07 AM
arccos
Quote:

Originally Posted by TheCoffeeMachine
$\displaystyle (1+ax+bx^2)^5 = \sum_{0\le k \le 5}\binom{5}{k}(ax+bx^2)^k = 1+5(ax+bx^2)+10(ax+bx^2)^2+...$.

By inspection, the coefficient of $x$ is $5a$ and that of $x^2$ is $10a^2+5b$, thus we have:

$5a = 10 \Rightarrow a = 2$ and $10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3$.

Oh yeah I could've grouped it another way... my bad. Haven't been doing expansions for a long time. Thanks!