Question #ce824

1 Answer
Oct 29, 2017

There is no solution

Explanation:

Using the Binomial Expansion, we have:

# (1+x)^n = ((n),(0)) + ((n),(1))x + ((n),(2))x^2 + ... + ((n),(n))x^n #

So, the coefficients of #x^4#, #x^5# and #x^6# are:

# ((n),(4))#, #((n),(5))# and # ((n),(6))#

respectively. So, Using the Combination definition:

# ((n),(r)) = (n!)/( r!(n-r)!)#

Then the coefficients are:

# Coeff(x^4): u_1 = (n!)/( 4!(n-4)!) #
# Coeff(x^5): u_2 = (n!)/( 5!(n-5)!) #
# Coeff(x^6): u_3 = (n!)/( 6!(n-6)!) #

Where it is assumed that #n gt 6# for the terms to actually exist. We require these term to be in GP #{u_1,u_2,u_3}={a, ar, ar^2}#, say. This would require:

# u_3/u_2 = u_2/u_1 => u_1 u_3 = u_2 ^2#

Hence, we have:

# (n!)/( 4!(n-4)!) * (n!)/( 6!(n-6)!) = (n!)/( 5!(n-5)!) * (n!)/( 5!(n-5)!) #

# :. 5!(n-5)! * 5!(n-5)! = 4!(n-4)! * 6!(n-6)! #

For clarity we perform a substitution:

# u = n-6 #

Then the expression becomes

# 5!(u+1)! 5!(u+1)! = 4!(u+2)! 6!u! #

We can then expand some of the factorials using #n! = n(n-1)!# to give:

# (5)4!(u+1)u! (5)(4!)(u+1)u! = 4!(u+2)(u+1)u! (6)(5)(4!)u! #

Many terms now cancel, leaving us with:

# 5(u+1) = 6(u+2) #
# :. 5u+5 = 6u+12 #
# :. 6u-5u = 5-12 #
# :. u = -7 #
# :. n-6 = -7 #
# :. n=-1 #

Our initial assertion was that #n gt 6# such that the binomial terms exist, and this criteria cannot be met to ensure the terms are in GP.

Hence there is no solution