How to find the coefficient of #x^50# in the expansion of the following?

Question

#(1+x)^1000 + x(1+x)^999 + x^2(1+x)^998 + . . . . +x^1000#

Options are:
A) #"^1000C_50#
B) #"^1001C_50#
C) #"^1002C_50#
D) None of the above

My work so far:
Since we want the term #x^50# only here, we get the term only till the following part in expansion given to us:
#(1+x)^1000 + x(1+x)^999 + . . . . +(1+x)^950x^50#
Now expanding the terms to single out those containing #x^50# using Binomial expansion I will get:
#"^1000C_50x^50# #+# #"^999C_49x^50# + #"^998C_48x^50# #+ . . . . +# #"^950C_0x^50#
#rArr# Sum of coefficients = #"^1000C_50# + #"^999C_49# + . . . +#"^950C_0#
I know the identity #"^nC_r# + #"^nC_(r+1)# = #"^(n+1)C_(r+1)#
But I couldn't use it to shorten the expansion.
How should I go about this?

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Answer 1 · 2018-06-26T13:25:47

#""^1001 C_50#

This is a GP of 1001 terms, with first term #(1+x)^1000# and common ratio #x/(1+x)# that can be easily summed up to give

#(1+x)^1001-x^1001#

It is easy to see from this that the desired coefficient( that of #x^50#) is #""^1001C_50#, since such a power can only come from the first term.