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Coefficient of x50 in (1 + x)1000 + 2x(1 + x)999 + 3x(1 + x)998 + .....is
  • a)
    1000C50
  • b)
    1001C50
  • c)
    1002C50
  • d)
    1002C49
Correct answer is option 'C'. Can you explain this answer?
Verified Answer
Coefficient of x50 in (1 + x)1000 + 2x(1 + x)999 + 3x(1 + x)998 + .......
Take (1 + x)1000 as common, after simplification it becomes (1 + x)1000 coefficient of x50 is 1002C50
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Most Upvoted Answer
Coefficient of x50 in (1 + x)1000 + 2x(1 + x)999 + 3x(1 + x)998 + .......
To find the coefficient of x^50 in the given expression, we need to identify the terms that contain x^50 and determine their coefficients.

Let's break down the given expression into individual terms:

Term 1: (1 x)^1000
Term 2: 2x(1 x)^999
Term 3: 3x(1 x)^998
...

The general term can be written as kx(1 x)^(1000-k), where k represents the coefficient of the term.

To find the coefficient of x^50, we need to solve the equation 1000 - k = 50, which gives us k = 950.

Now, let's calculate the coefficient of x^50 in the first term:

Coefficient of x^50 in (1 x)^1000 = (1000C50) = 1000! / (50! * 950!)

Similarly, the coefficient of x^50 in the second term is:

Coefficient of x^50 in 2x(1 x)^999 = 2 * (999C49) = 2 * 999! / (49! * 950!)

And for the third term:

Coefficient of x^50 in 3x(1 x)^998 = 3 * (998C48) = 3 * 998! / (48! * 950!)

We can observe that the coefficient of x^50 increases by 1 for each subsequent term.

Therefore, the coefficient of x^50 in the given expression can be calculated as:

1000! / (50! * 950!) + 2 * 999! / (49! * 950!) + 3 * 998! / (48! * 950!) + ...

This can be simplified by factoring out 950! from each term:

950! * [1000/(50! * 950!) + 2 * 999/(49! * 950!) + 3 * 998/(48! * 950!) + ...]

Simplifying further, we get:

950! * [1000/(50!) + 999/(49!) + 998/(48!) + ...]

This can be written as:

950! * [Summation of (1000-k)/(k!) from k = 0 to 950]

Using the formula for the sum of a series, we can rewrite it as:

950! * [1001C51]

Therefore, the coefficient of x^50 in the given expression is 1001C51 or option 'C'.
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