Question:
The coefficient of \(x^{99}\) in the expansion of \((x-1)(x-2)(x-3)\dots(x-100)\) is:
\((x-1)(x-2)(x-3)\dots(x-100)\) के विस्तार में \(x^{99}\) का गुणांक है:
Options:
- 5050
- 5000
- -5050
- -5000
- 5050
- 5000
- -5050
- -5000
Correct Answer: c)
Explanation:
In a polynomial product \((x – a_1)(x – a_2)\dots(x – a_n)\), the coefficient of \(x^{n-1}\) is equal to \(-(a_1 + a_2 + \dots + a_n)\).
Here, \(n = 100\) and \(a_k = k\) for \(k = 1, 2, \dots, 100\).
\[\text{Coefficient of } x^{99} = -(1 + 2 + 3 + \dots + 100) = -\frac{100 \times 101}{2} = -5050\]
Hence, the correct option is (c).
बहुपद गुणनफल \((x – a_1)(x – a_2)\dots(x – a_n)\) में \(x^{n-1}\) का गुणांक \(-(a_1 + a_2 + \dots + a_n)\) होता है।
यहाँ \(n = 100\) तथा \(a_k = k\) है:
\[\text{गुणांक} = -(1 + 2 + 3 + \dots + 100) = -\frac{100 \times 101}{2} = -5050\]
अतः सही विकल्प (c) है।