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

## ${\left(1 + x\right)}^{1000} + x {\left(1 + x\right)}^{999} + {x}^{2} {\left(1 + x\right)}^{998} + . . . . + {x}^{1000}$ 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: ${\left(1 + x\right)}^{1000} + x {\left(1 + x\right)}^{999} + . . . . + {\left(1 + x\right)}^{950} {x}^{50}$ 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 $\Rightarrow$ 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?

""^1001 C_50
This is a GP of 1001 terms, with first term ${\left(1 + x\right)}^{1000}$ and common ratio $\frac{x}{1 + x}$ that can be easily summed up to give
${\left(1 + x\right)}^{1001} - {x}^{1001}$
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.