A binomial expansion problem I am having problem with.

Question

When $ (1+cx)^n$ is expanded a series in ascending powers of $x$, the first three terms are given by $1+20x+150x^2$. Calculate the value of the constants $c$ and $n$.

If possible please include working so I could perhaps understand how you got to the answer.

Answer 1

You need to use the Binomial formula

Applying it you get:

$$(1+cx)^n = {n \choose 0} \cdot 1^n \cdot (cx)^{0} + {n \choose 1} \cdot 1^{n-1} \cdot (cx)^{1} + {n \choose 2} \cdot 1^{n-2} \cdot (cx)^{2} +\ ... $$

$$(1+cx)^n = {n \choose 0} + {n \choose 1} \cdot (cx)^{1} + {n \choose 2} \cdot (cx)^{2} +\ ... $$

$$(1+cx)^n = {n \choose 0} + {n \choose 1} \cdot cx + {n \choose 2} \cdot c^2x^{2} +\ ... $$

Now you can construct equations for $c,n$ because you are given the coefficients before $x$ and $x^2$.

$${n \choose 1} \cdot c = 20 $$ $${n \choose 2} \cdot c^2 = 150 $$

Let us solve this system of two equations.

$$nc= 20$$ $$\frac{n(n-1)}{2}c^2 = 150$$

We get

$$nc = 20$$ $$(n-1)c = 15$$

This means

$$nc = 20$$ $$nc-c = 15$$

This gives us the answer which is $c=5, n=4$

Answer 2

$$(1+cx)^n=1+ncx+\frac{1}{2}n(n-1)(cx)^2+...$$ So we have $nc=20~~~(1)$ and $\frac{1}{2}n(n-1)c^2 =150~~~(2)$ $$\implies n(n-1)c^2=300 \implies 20 (n-1)c=300 \implies (n-1)C=15 \implies 20-c=15 $$ $$\implies c=5, n=4$$