# Thread: [SOLVED] Binomial Theorem help...

1. ## [SOLVED] Binomial Theorem help...

Question:
Given that the expansion of $(1+ax)^n$ begins $1+36x+576x^2$, find the values of $a$ and $n$.

2. Originally Posted by looi76
Question:
Given that the expansion of $(1+ax)^n$ begins $1+36x+576x^2$, find the values of $a$ and $n$.
$36x = nax \implies 36 = na$.
$575x^2 = \frac{n(n-1)}{2} a^2 x^2 \implies 575 = \frac{n(n-1)}{2} a^2$
Now solve these equations.

3. Originally Posted by looi76
Question:
Given that the expansion of $(1+ax)^n$ begins $1+36x+576x^2$, find the values of $a$ and $n$.
Well
$(1 + ax)^n = \sum_{k = 0}^n {n \choose k}(ax)^k = 1 + n \cdot ax + \frac{n(n - 1)}{2} \cdot a^2x^2 + ~...$

Thus
$na = 36$
and
$\frac{n(n - 1)}{2} \cdot a^2 = 576$

So
$a = \frac{36}{n}$

giving:
$\frac{n(n - 1)}{2} \cdot \left ( \frac{36}{n} \right ) ^2 = 576$

$\frac{n - 1}{2n} \cdot 1296 = 576$

Giving me n = 9 and a = 4.

-Dan