The first four terms in binomial expansion of (a+b)ⁿ , in descensing power of a , are w,x, y and z respectively.Show that (n-2)=3nwz?

1 Answer
Steve M
May 28, 2018

If we recall the Binomial Theorem, we have:

# (a+b)^n = sum_(k=0)^n \ ( ( n), (k) ) \ a^(n-k) \ b^k #

Where:

# ( (n), (k) ) = ""_nC^r = (n!)/((n-r)!r!) #

is the binomial (or combinatorial coefficient. So, if we expand the first four terms, we have:

# (a+b)^n = ( (n), (0) ) a^(n) + ( (n), (1) ) a^(n-1)b^1 + ( (n), (2) ) a^(n-2)b^2 #
# \ \ \ \ \ \ \ \ \ \ \ \ \ \ + ( (n), (3) ) a^(n-3)b^3 + ...#

So given the coefficient terms are #w,x,y# and #z#, we can write:

# w = ( (n), (0) ) #, # x = ( (n), (1) ) #, # y = ( (n), (2) ) #, # z=( (n), (3) ) #

So consider the product, given by the RHS of the given expression:

# RHS = 3nwz #

# \ \ \ \ \ \ \ \ = 3n \ ( (n), (0) ) \ ( (n), (3) ) #

# \ \ \ \ \ \ \ \ = 3n \ (n!)/((n-0)!0!) \ (n!)/((n-3)!3!) #

# \ \ \ \ \ \ \ \ = 3n \ (n!)/(n!) \ (n(n-1)(n-2)(n-3)!)/((n-3)! \ 6) #

# \ \ \ \ \ \ \ \ = 1/2n \ n(n-1)(n-2) #

# \ \ \ \ \ \ \ \ = 1/2n^2(n-1)(n-2) #, contrary to the given result

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