Solution

Easiest Method — Coefficient of \( x^{50} \)

Given Expression:

\[ (1 + x)^{1000} + 2x(1 + x)^{999} + 3x^2(1 + x)^{998} + \cdots + 1001x^{1000} \]

This follows a known identity that simplifies the full expression to:

\[ f(x) = (1 + x)^{1002} \]

Now: The coefficient of \( x^{50} \) in \( f(x) \) is:

\[ \boxed{\binom{1002}{50}} \]

✅ Final Answer:   \( \boxed{\binom{1002}{50}} \)

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Solution

First expression is:

$\left(x^2+\frac{1}{x}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x^2)^{12-k}\left(\frac{1}{x}\right)^k$

$T_{k+1}={}^{12}C_k x^{24-2k-k}$

$T_{k+1}={}^{12}C_k x^{24-3k}$

For coefficient of $x^{10}$,

$24-3k=10$

$3k=14$

$k=\frac{14}{3}$

This is not an integer, so $x^{10}$ term is not present in the first expression.

Now second expression is:

$\left(x+\frac{1}{x^2}\right)^{12}$

Its general term is:

$T_{k+1}={}^{12}C_k(x)^{12-k}\left(\frac{1}{x^2}\right)^k$

$T_{k+1}={}^{12}C_k x^{12-k-2k}$

$T_{k+1}={}^{12}C_k x^{12-3k}$

For coefficient of $x^{10}$,

$12-3k=10$

$3k=2$

$k=\frac{2}{3}$

This is also not an integer, so $x^{10}$ term is not present in the second expression.

Therefore, the coefficient of $x^{10}$ is $0$.

Solution