Question

Question #e11e3

Answer 1

`n=8`

Explanation:

We know `(a+b)^n= nC_0 a^n*b^0 +nC_1 a^(n-1)*b^1 + nC_2 a^(n-2)*b^2+..........+nC_n a^(n-n)*b^n`

Here `a=x,b=1,` We know, `nC_r = (n!)/(r!*(n-r)!`

`(x+1)^n` .Coefficient of `x^3` is `C_(x^3)=nC_(n-3)*(1)^(n-3)` or

`C= (n!)/((n-3)!*3!)*1 = ((n)(n-1)(n-2))/6`

Coefficient of `x^2` is `C_(x^2)=nC_(n-2)*(1)^(n-2)` or

`C= (n!)/((n-2)!*2!)*1 = ((n)(n-1))/2` .By given condition ,

`C_(x^3)=2*C_(x^2)` or

`(cancel((n)(n-1))(n-2))/6=cancel2*cancel((n)(n-1))/cancel2` or

`(n-2)/6=1 or (n-2)=6 or n=8` [Ans]