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

First you line up all the numbers. Then you cross off each number end from end. Whatever number is the final number uncrossed that is your answer. If it is 2 simply add both of them up and divide by 2