Question:

Step 1: Determine $f(n)$. The given expression is $(1+x)^{2n} + (1+2x+x^2)^n$. Recognize that $1+2x+x^2 = (1+x)^2$. Thus, the expression can be rewritten as: $$ (1+x)^{2n} + ((1+x)^2)^n = (1+x)^{2n} + (1+x)^{2n} = 2(1+x)^{2n} $$ The sum $\sum_{r=0}^{2n} a_r$ is obtained by substituting $x=1$ into the polynomial. Therefore, $f(n)$ is given by: $$ f(n) = 2(1+1)^{2n} = 2 \cdot 2^{2n} = 2^{2n+1} $$ Step 2: Compute $\sum_{n=1}^{\infty} \dfrac{1}{f(n)}$. Substitute $f(n) = 2^{2n+1}$ into the sum: $$ \sum_{n=1}^{\infty} \frac{1}{f(n)} = \sum_{n=1}^{\infty} \frac{1}{2^{2n+1}} $$ This can be rewritten as: $$ \sum_{n=1}^{\infty} \frac{1}{2 \cdot (2^2)^n} = \sum_{n=1}^{\infty} \frac{1}{2 \cdot 4^n} = \frac{1}{2} \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n $$ This is an infinite geometric series with the first term $a = \frac{1}{4}$ (for $n=1$) and common ratio $r = \frac{1}{4}$. The sum of an infinite geometric series is $\frac{a}{1-r}$. $$ \sum_{n=1}^{\infty} \left(\frac{1}{4}\right)^n = \frac{\frac{1}{4}}{1 - \frac{1}{4}} = \frac{\frac{1}{4}}{\frac{3}{4}} = \frac{1}{3} $$ Therefore, the sum is: $$ \sum_{n=1}^{\infty} \frac{1}{f(n)} = \frac{1}{2} \cdot \frac{1}{3} = \frac{1}{6} $$ Step 3: Determine the largest value of $p$ for which $f(5)$ is divisible by $2^p$. Substitute $n=5$ into the expression for $f(n)$: $$ f(5) = 2^{2(5)+1} = 2^{10+1} = 2^{11} $$ For $f(5)$ to be divisible by $2^p$, $2^p$ must be a factor of $2^{11}$. The largest integer value of $p$ for which $2^p$ divides $2^{11}$ is $p=11$.