Question

The coefficient of x\(^{48}\) in \(1(1+x)+2(1+x)^2+3(1+x)^3 +.....+100(1+x)^{100}\) is:

• A. \(^{100}\)C\(_{50}\) \(^{101}\)C\(_{50}\) – \(^{101}\)C\(_{49}\)
• B. 100\(^{101}\)C\(_{49}\) – \(^{101}\)C\(_{50}\)
• C. \(^{101}\)C\(_{46}\) – 100
• D. \(^{101}\)C\(_{47}\) – \(^{101}\)C\(_{46}\)

Hint:

Finding coefficients in complex series can often be simplified by first finding a closed-form expression for the sum of the series. For AGP, the method of subtracting a multiple of the series from itself is standard.

Answer 1

The Correct Option is B

Step 1: Identify the Series:
Let the given series be S. This is an Arithmetico-Geometric Progression (AGP). Let \(y = 1+x\). The series can be written as: \[ S = 1 \cdot y + 2 \cdot y^2 + 3 \cdot y^3 + \dots + 100 \cdot y^{100} \]
Step 2: Summing the AGP:
To find the sum of this AGP, we use the standard method: \[ S = y + 2y^2 + 3y^3 + \dots + 100y^{100} \quad (1) \] Multiply by y: \[ yS = y^2 + 2y^3 + \dots + 99y^{100} + 100y^{101} \quad (2) \] Subtracting (2) from (1): \[ S(1-y) = (y + y^2 + y^3 + \dots + y^{100}) - 100y^{101} \] The terms in the parenthesis form a Geometric Progression (GP) with first term y, common ratio y, and 100 terms. The sum of this GP is \(\frac{y(y^{100}-1)}{y-1}\). \[ S(1-y) = \frac{y(y^{100}-1)}{y-1} - 100y^{101} \]
Step 3: Express S in terms of x:
Substitute \(y=1+x\) back into the equation. Note that \(1-y = -x\) and \(y-1 = x\). \[ S(-x) = \frac{(1+x)((1+x)^{100}-1)}{x} - 100(1+x)^{101} \] \[ S(-x) = \frac{(1+x)^{101} - (1+x)}{x} - 100(1+x)^{101} \] Multiply by \(-1/x\) to solve for S: \[ S = -\frac{(1+x)^{101} - (1+x)}{x^2} + \frac{100(1+x)^{101}}{x} \] \[ S = \frac{100(1+x)^{101}}{x} - \frac{(1+x)^{101}}{x^2} + \frac{1+x}{x^2} \]
Step 4: Find the Coefficient of x\(^{48}\):
We need to find the coefficient of \(x^{48}\) in the expression for S. Let's look at each term.
For the term \(\frac{100(1+x)^{101}}{x}\), we need the coefficient of \(x^{49}\) in the expansion of \(100(1+x)^{101}\). This is \(100 \cdot ^{101}C_{49}\).
For the term \(-\frac{(1+x)^{101}}{x^2}\), we need the coefficient of \(x^{50}\) in the expansion of \(-(1+x)^{101}\). This is \(-^{101}C_{50}\).
The term \(\frac{1+x}{x^2} = \frac{1}{x^2} + \frac{1}{x}\) has no terms with non-negative powers of x, so it doesn't contribute to the coefficient of \(x^{48}\). Combining the contributions, the coefficient of \(x^{48}\) in S is: \[ 100 \cdot ^{101}C_{49} - ^{101}C_{50} \]
Step 5: Final Answer:
The coefficient of x\(^{48}\) is \(100 \cdot ^{101}C_{49} - ^{101}C_{50}\).