1. Triple term binomial expansion

In the expansion $\displaystyle (1 + ax + bx^2)^5$ the coefficients of $\displaystyle x$ and $\displaystyle x^2$ 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 $\displaystyle ((1 + ax) + bx^2)^5$ and prepare it for binomial expansion but I am getting something like
$\displaystyle \displaystyle 5 \choose 0$ $\displaystyle \displaystyle (1+ax)^5(bx^2)^0 + $$\displaystyle \displaystyle 5 \choose 1$$\displaystyle \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.

2. Use:

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

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

Fernando Revilla

3. the trinomial theorem is not in my syllabus. is there another way around it?

4. 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

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

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

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

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

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

$\displaystyle 5a = 10 \Rightarrow a = 2$ and $\displaystyle 10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3$.
You should point out that the ellipsis ... contains only terms of order greater than 2.

CB

7. Originally Posted by CaptainBlack
You should point out that the ellipsis ... contains only terms of order greater than 2.
I should have, indeed. Thanks.

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

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

$\displaystyle 5a = 10 \Rightarrow a = 2$ and $\displaystyle 10a^2+b = 25 \Rightarrow 40+5b = 25 \Rightarrow b = -3$.
Oh yeah I could've grouped it another way... my bad. Haven't been doing expansions for a long time. Thanks!