In the expansion of (1+2x)^n, the coefficient of x^4 is ten times the coefficient of x^2. find the value of the positive integer n.

Hello,

Coeff of x² is C(n,2)*2²=n(n-1)/2* 4=2n(n-1)

Coeff of x^4 is 2^4*C(n,4)=16*n(n-1)(n-2)(n-3)/ 4!=2*n(n-1)(n-2)(n-3)/3

So,
2*n(n-1)(n-2)(n-3)/3=10*2n(n-1)
==>(n-2)(n-3)=30
==>n²-5n-24=0 (delta= 25+4*24=11²)
==>n=(5+11)/2=8 or n=(5-11)/2=-3 (excluded)

==>(1+2x)^8=256x^8+1024x^7+1792x^6+1792x^5+1120x^4+448x^3+112x^2+2x+1
and 1120=10*112

Y’all got to do it all over