Question

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.

Answer 1

Recall the binomial theorem:

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

where

$\dbinom nk=\dfrac{n!}{k!(n-k)!}$

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

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

The x³ occurs for k = 3, with coefficient

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

The latter coefficient is twice the other one, so that

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

Solve for n :

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

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

$n-2=6$

$\boxed{n=8}$

Answer 2

According to Khayyam-Pascal triangle ,

the value of n is 8.

$look \: \\ \\ {(x + 1)}^{8} = {x}^{8} + 8 {x}^{7} + 28 {x}^{6} + 56 {x}^{5} + 70 {x}^{4} + 56 {x}^{3} + 28 {x}^{2} + 8x + 1 \\ \frac{56}{28} = 2$