Binomial Coefficient Identification
The coefficient of $ x^{50} $ in
\[(1 + x)^{1000} + 2x (1 + x)^{999} + 3x^2 (1 + x)^{998} + \dots + 1001x^{1000}\]can be expressed in the form $ \binom{n}{k} $. Find the smallest possible value of $ n + k $.
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- $\frac{a}{b}$
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- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
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- $\cap$
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- $\infty$
Solution
Let
\[S = (1 + x)^{1000} + 2x (1 + x)^{999} + \dots + 1000x^{999} (1 + x) + 1001x^{1000}\]Then
\begin{align*}
xS &= x (1 + x)^{1000} + 2x^2 (1 + x)^{999} + \dots + 1000x^{1000} (1 + x) + 1001x^{1001}, \\
(1 + x)S &= (1 + x)^{1001} + 2x (1 + x)^{1000} + \dots + 1000x^{999} (1 + x)^2 + 1001x^{1000} (1 + x).\end{align*}Subtracting these equations, we get
\[S = (1 + x)^{1001} + x(1 + x)^{1000} + \dots + x^{999} (1 + x)^2 + x^{1000} (1 + x) - 1001x^{1001}.\]Then
\begin{align*}
xS &= x(1 + x)^{1001} + x^2 (1 + x)^{1000} + \dots + x^{1000} (1 + x)^2 + x^{1001} (1 + x) - 1001x^{1002}, \\
(1 + x)S &= (1 + x)^{1002} + x (1 + x)^{1001} + \dots + x^{999} (1 + x)^3 + x^{1000} (1 + x)^2 - 1001x^{1001} (1 + x).\end{align*}Subtracting these equations, we get
\[S = (1 + x)^{1002} - 1002x^{1001} (1 + x) + 1001x^{1002}.\]By the Binomial Theorem, the coefficient of $ x^{50} $ is then $ \binom{1002}{50} $. The final answer is $ 1002 + 50 = \boxed{1052} $.
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