If the coefficient of \( x^r \) in the expansion of \( (1 + x + x^2)^{100} \) is \( a_r \), and \( S = \sum\limits_{r=0}^{300} a_r \), then
\[ \sum\limits_{r=0}^{300} r a_r = \]
\( (100) S \)
Step 1: Understanding the Given Expansion
The given function is: \[ f(x) = (1 + x + x^2)^{100}. \] The coefficient of \( x^r \) in the expansion of this expression is denoted as \( a_r \), meaning: \[ S = \sum\limits_{r=0}^{300} a_r = f(1). \]
Step 2: Finding \( S \)
To determine \( S \), we evaluate the function at \( x = 1 \): \[ S = f(1) = (1 + 1 + 1)^{100} = 3^{100}. \]
Step 3: Computing \( \sum r a_r \)
By differentiation, \[ f'(x) = 100(1 + x + x^2)^{99} \cdot (1 + 2x). \] Setting \( x = 1 \), \[ \sum\limits_{r=0}^{300} r a_r = f'(1). \] Substituting \( x = 1 \) into \( f'(x) \): \[ f'(1) = 100 (3^{99}) \cdot (1 + 2) = 100 \times 3^{99} \times 3 = 300 \times 3^{99}. \] Thus, \[ \sum\limits_{r=0}^{300} r a_r = 150 S. \]
Given below are two statements, one is labelled as Assertion (A) and the other one labelled as Reason (R).
Assertion (A): \[ 1 + \frac{2.1}{3.2} + \frac{2.5.1}{3.6.4} + \frac{2.5.8.1}{3.6.9.8} + \dots \infty = \sqrt{4} \] Reason (R): \[ |x| <1, \quad (1 - x)^{-1} = 1 + nx + \frac{n(n+1)}{1.2} x^2 + \frac{n(n+1)(n+2)}{1.2.3} x^3 + \dots \]
\[ \text{The domain of the real-valued function } f(x) = \sin^{-1} \left( \log_2 \left( \frac{x^2}{2} \right) \right) \text{ is} \]
Let \( A = [a_{ij}] \) be a \( 3 \times 3 \) matrix with positive integers as its elements. The elements of \( A \) are such that the sum of all the elements of each row is equal to 6, and \( a_{22} = 2 \).
\[ \textbf{If } | \text{Adj} \ A | = x \text{ and } | \text{Adj} \ B | = y, \text{ then } \left( | \text{Adj}(AB) | \right)^{-1} \text{ is } \]