Question

Find the coefficient of x⁵⁰ after simplifying and collecting the like terms in the expansion of

a. (1+x)¹⁰⁰⁰+x(1+x)⁹⁹⁹+x2(1+x)⁹⁹⁸++x¹⁰⁰⁰

Answer 1

The coefficient of
`\(x^(50)\)` in the given expression
`\((1+x)^(1000)+x(1+x)^(999)+x^2(1+x)^(998)+\ldots+x^(1000)\)` is the sum of binomial coefficients from C(950, 50) to C(1000, 0).

The given expression is a part of the binomial expansion, specifically for the binomial (1 + x)^1000. To find the coefficient of
`\(x^(50)\)`, you need to consider the terms that contribute to this power when the expression is expanded.

In the expansion of
`\((1 + x)^n\)`, the general term is given by:

`\[ C(n, k) \cdot (x^k) \cdot (1^(n-k)) \]`

where C(n, k) is the binomial coefficient, which is the number of ways to choose k elements from a set of n distinct elements.

In the given expression, we have terms of the form x^n where n ranges from 1000 to 100, and each term is multiplied by a power of x.

The term that contributes to
`\(x^(50)\)` comes from the term
`\(x^(50)\)` in the expansion of
`\((1 + x)^(950)\)` in the first term,
`\(x^(49)\)` in the expansion of
`\((1 + x)^(951)\)` in the second term, and so on.

The coefficient of
`\(x^(50)\)` is the sum of the coefficients of these terms. Therefore, the coefficient of
`\(x^(50)\)` in the given expression is:

`\[ C(950, 50) + C(951, 49) + C(952, 48) + \ldots + C(1000, 0) \]`

You can use a binomial coefficient calculator or a mathematical software to compute this sum. It's a cumbersome calculation, but the final result is the coefficient of
`\(x^(50)\)` in the expansion of the given expression.