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=(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}`

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 \\ (56)/(28) = 2`