Chapter 9.4, Problem 2QCE

Given that $a_1=3,{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{a}_k}=1,\text{and}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}{b}_k}=5$ it follows that ${\displaystyle \underset{k=2}{\overset{\infty }{\sum }}{a}_k}=\text{\_\_\_\_\_\_}\text{and}{\displaystyle \underset{k=1}{\overset{\infty }{\sum }}\left(2a_k+b_k\right)}=\text{\_\_\_\_\_\_}$