Something my AP Calculus class does not teach: Advanced Binomial?

Since $a_k = \binom{n}{k}2^k$ for $0 \le k \le n$ we have $\begin{aligned} \sum_{k=0}^n (3k+1)a_k & = \; 3\sum_{k=1}^n k\binom{n}{k} 2^k + (2 + 1)^n \; = \; 3n\sum_{k=1}^n \binom{n-1}{k-1}2^k + 3^n \\ & = \; 3n\sum_{k=0}^{n-1}\binom{n-1}{k}2^{k+1} + 3^n \; = \; 6n(2 + 1)^{n-1} + 3^n \; = \; (2n+1)3^n \end{aligned}$ making the answer $2+1+3=\boxed{6}$ .

Starting with ${ \left( 1+2x \right) }^{ n }=\sum _{ k=0 }^{ n }{ { a }_{ k }{ x }^{ k } }$ Putting $x=1$ , we have: $\sum _{ k=0 }^{ n }{ { a }_{ k } } ={ 3 }^{ n }$

Differentiating both sides of ${ \left( 1+2x \right) }^{ n }=\sum _{ k=0 }^{ n }{ { a }_{ k }{ x }^{ k } }$ w.r.t. x, we have: $2n{ \left( 1+2x \right) }^{ n-1 }=\sum _{ k=1 }^{ n }{ k{ a }_{ k }{ x }^{ k-1 } }$

Note that the term with $k=0$ will vanish after differentiation.

Putting $x=1$ again, we have: $\sum _{ k=0 }^{ n }{ k{ a }_{ k } } =2n{ 3 }^{ n-1 }$

Notice that $\sum _{ k=0 }^{ n }{ \left( 3k+1 \right) { a }_{ k } } =3\sum _{ k=0 }^{ n }{ k{ a }_{ k } } +\sum _{ k=0 }^{ n }{ { a }_{ k } } =3\sum _{ k=1 }^{ n }{ k{ a }_{ k } } +\sum _{ k=0 }^{ n }{ { a }_{ k } }$

Therefore, $\sum _{ k=0 }^{ n }{ \left( 3k+1 \right) { a }_{ k } } =3\left( 2n{ 3 }^{ n-1 } \right) +{ 3 }^{ n }=2n{ 3 }^{ n }+{ 3 }^{ n }=\left( 2n+1 \right) { 3 }^{ n }$

So, $p=2$ , $q=1$ , $r=3$ , and $p+q+r=2+1+3=\boxed 6$