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

(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?

1 Answer
Jun 26, 2018

""^1001 C_50

Explanation:

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.