m and n positive integers, m‹n.expansion of (1+2x)power n, coefficient of x power m is 3/2 times coefficient of x power m-1,show that m and n satisfy 4n-7m+4=0.find smallest values of m and n

Question

m‹n is m is less than and equal to n

Answer 1

(1+2x)^n=1+2nx+n(n-1)/2*4x^2+n(n-1)(n-2)/6*8x^3+...+n(n-1)(...)(n-m+1)/m!(2x)^m (up to mth term).

If we take two consecutive terms m-1 and m, a[m-1]x^(m-1) and a[m]x^m, we can relate the terms:

mth term: n(n-1)(...)(n-m+1)/m!(2x)^m

(m-1)th term: n(n-1)(...)(n-m)/(m-1)!(2x)^(m-1)

mth term/(m-1)th term=(n-m+1)/m * 2x=3/2.

(2x)^m=2^m*x^m and (2x)^(m-1)=2^(m-1)*x^(m-1). So there is a factor of 2 between two consecutive powers of x.

Therefore 2(n-m+1)/m=3/2; 4n-4m+4=3m; 4n-7m+4=0 QED.

Therefore m=(4n+4)/7 so 4n+4=4(n+1) has to be a multiple of 7, i.e., n+1 must be a multiple of 7, so n=6 is the lowest value, and m=4.

CHECK Coefficients when n=6 are 1 6 15 20 15 6 1 (coefficients of (2x)), and m starts from 0 because (2x)^0=1, the first term.

When we include the factor of 2 we get: 1 12 60 160 240 192 64. When m=4 we have 160 and 240 as consecutive terms and 240/160=3/2.