Question:

Step 1: Determine the general term of the sequence $\{a_n\}$. The sequence $\{a_n\}$ is an arithmetic progression with common difference $d=2$, since $a_{n+1} = 2+a_n$. We are given $a_{10}=21$. The general term for an arithmetic progression is $a_n = a_1 + (n-1)d$. Substituting the given values: $a_{10} = a_1 + (10-1)2$ $21 = a_1 + 9 \cdot 2$ $21 = a_1 + 18$ $a_1 = 3$ Thus, the general term for $a_n$ is: $a_n = 3 + (n-1)2 = 3 + 2n - 2 = 2n+1$. Step 2: Express $b_n$ using partial fraction decomposition. The sequence $b_n$ is defined as $b_n = \dfrac{1}{a_n a_{n+2}}$. Using $a_n = 2n+1$: $a_{n+2} = 2(n+2)+1 = 2n+4+1 = 2n+5$. So, $b_n = \dfrac{1}{(2n+1)(2n+5)}$. We decompose $b_n$ into partial fractions: $\dfrac{1}{(2n+1)(2n+5)} = \dfrac{A}{2n+1} + \dfrac{B}{2n+5}$ Multiplying by $(2n+1)(2n+5)$ gives: $1 = A(2n+5) + B(2n+1)$ Setting $2n+1=0 \implies n=-1/2$: $1 = A(2(-1/2)+5) + B(0) \implies 1 = A(4) \implies A = \dfrac{1}{4}$. Setting $2n+5=0 \implies n=-5/2$: $1 = A(0) + B(2(-5/2)+1) \implies 1 = B(-4) \implies B = -\dfrac{1}{4}$. Therefore, $b_n = \dfrac{1}{4(2n+1)} - \dfrac{1}{4(2n+5)} = \dfrac{1}{4}\left(\dfrac{1}{2n+1} - \dfrac{1}{2n+5}\right)$. Step 3: Calculate the sum of the series $\sum_{r=1}^{\infty}b_r$. This is a telescoping series. Let $S_N = \sum_{r=1}^{N} b_r$. $S_N = \sum_{r=1}^{N} \dfrac{1}{4}\left(\dfrac{1}{2r+1} - \dfrac{1}{2r+5}\right)$ $S_N = \dfrac{1}{4} \left[ \left(\dfrac{1}{3} - \dfrac{1}{7}\right) + \left(\dfrac{1}{5} - \dfrac{1}{9}\right) + \left(\dfrac{1}{7} - \dfrac{1}{11}\right) + \left(\dfrac{1}{9} - \dfrac{1}{13}\right) + \dots + \left(\dfrac{1}{2N-1} - \dfrac{1}{2N+3}\right) + \left(\dfrac{1}{2N+1} - \dfrac{1}{2N+5}\right) \right]$ Notice that the terms cancel out in pairs, with a shift of two terms. For example, $-\frac{1}{7}$ from $r=1$ cancels with $+\frac{1}{7}$ from $r=3$. The terms that remain are the first two positive terms and the last two negative terms: $S_N = \dfrac{1}{4} \left[ \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{2N+3} - \dfrac{1}{2N+5} \right]$ Now, we find the sum to infinity by taking the limit as $N \to \infty$: $\sum_{r=1}^{\infty} b_r = \lim_{N\to\infty} S_N = \lim_{N\to\infty} \dfrac{1}{4} \left[ \dfrac{1}{3} + \dfrac{1}{5} - \dfrac{1}{2N+3} - \dfrac{1}{2N+5} \right]$ As $N \to \infty$, $\dfrac{1}{2N+3} \to 0$ and $\dfrac{1}{2N+5} \to 0$. So, $\sum_{r=1}^{\infty} b_r = \dfrac{1}{4} \left[ \dfrac{1}{3} + \dfrac{1}{5} \right]$ $\sum_{r=1}^{\infty} b_r = \dfrac{1}{4} \left[ \dfrac{5+3}{15} \right] = \dfrac{1}{4} \left[ \dfrac{8}{15} \right] = \dfrac{2}{15}$. Step 4: Calculate the value of $15\sum_{r=1}^{\infty}b_r$. $15 \sum_{r=1}^{\infty}b_r = 15 \cdot \dfrac{2}{15} = 2$.