Question #ce824
1 Answer
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
# ((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
# 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
# (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
Hence there is no solution