In the expression of (x + 1)^n, the coefficient of the x^3 term is two times the coefficient of the x^2 term. Find the value of n.

Question

asked Jul 15, 2021 215k views

2 Answers

Teka · Answer 1 · 2021-07-17T12:49:42+0000

Recall the binomial theorem:

$(a+b)^n=\displaystyle\sum_(k=0)^n\binom nk a^k b^(n-k)$

where

$\dbinom nk=(n!)/(k!(n-k)!)$

is the binomial coefficient. So

$(x+1)^n=\displaystyle\sum_(k=0)^n\binom nk x^k$

The x² term occurs for k = 2, which has coefficient

$\dbinom n2=(n!)/(2!(n-2)!)=\frac{n(n-1)}2$

The x³ occurs for k = 3, with coefficient

$\dbinom n3=(n!)/(3!(n-3)!)=\frac{n(n-1)(n-2)}6$

The latter coefficient is twice the other one, so that

$\frac{n(n-1)(n-2)}6=2\cdot\frac{n(n-1)}2$

Solve for n :

$\frac{n(n-1)(n-2)}6=n(n-1)$

n(n-1)(n-2)=6n(n-1)

n-2=6

$\boxed{n=8}$