Finding the power of a binomial

Question

Question: The first three terms in the expansion of $(1+ax)^n$ are $1+35x+490x^2$. Given that n is a positive integer, find the value of n and a.

So I have tried to construct a general equation $1+anx+\frac{n!}{(n-2)!2!}(ax)^2$ because using pascals triange, I know that the coefficient of the second term will be n. For the third term I have used the formula $nCr\equiv \frac{n!}{(n-r)!r!}$

Which means that $an=35$ and $\frac{n!}{(n-2)!2!}a^2=490$. I thought maybe from here I could solve them simultaneously by substituting $\frac{35}{a}$ into the other equation, although I cant rearrange due to the factorials.

Is there some way of simplifying the factorials or have I done something wrong?

The answer is $n=5$ and $a=7$

Answer 1

we get $$a=\frac{35}{n}$$ and so $$\frac{n(n-1)}{2}\cdot \frac{35^2}{n^2}=490$$ and form here we obtain $$1225n-1225=980n$$ Can you proceed?

Answer 2

According to the binomial theorem $$(1+ax)^n=1+n(ax)+\dfrac{n(n-1)}{2}(ax)^2+\cdots.$$ Now solve $na=35$ and $n(n-1)a^2=980$ simultaneously.

Answer 3

You have the equations $$ na=35,\qquad \frac{n(n-1)}2\,a^2=980. $$ Rewrite the second equation as $$ (na)^2-(na)a = 1960\iff a=na-\frac{1960}{na}=35 -28=7, $$ so that $\;n=\dfrac{35}7=5$.

Answer 4

You can also work with derivatives:

$$((1+ax)^n)'=na(1+ax)^{n-1},$$

$$((1+ax)^n)''=n(n-1)a^2(1+ax)^{n-2}.$$

Then setting $x=0$ and comparing to the derivatives of the expansion,

$$na=35,\\n(n-1)a^2=2\cdot490=35(n-1)a$$ and by elimination of $n$,

$$a=7, \\n=5.$$