Question

The coefficient of x^2 in the expansion of (1+2x)^n is 60. given that n>0, find the value of n.​

Answer 1

n=6

Answer:

Solution given:

The coefficient of x^2=60

we have,

$(a+x)^n=C(n,0)a^n+C(n,1)a^(n-1)x+C(n,2)a^(n-2)x^2+C(n,3)a^(n-3)x^3....C(n,r)a^(n-r)x^r+C(n,n)x^n$

we get x² in 3rd term,so

3rd term of  (1+2x)^n is C(n,2)*1^n-2)(2x)^2=C(n,2)4x²

since 1^n is 1.

we have a coefficient of x^2 is 60, so

C(n,2)4=60

C(n,2)=60/4

$\frac{n!}{(n-2)!*2!}$=15

$\frac{n(n-1)(n-2)!}{(n-2)!*2!}$=15

n(n-1)=15*2

n^2-n=30

n^2-n-30=0

doing middle term factorization

n^2-6x+5x-30=0

n(n-6)+5(n-6)=0

(n-6)(n+5)=0

either n-6=0

n=6

or,

n+5=0

n=-5 since n>0

so neglected.

So the value of n is 6.

Step-by-step explanation: