Let $b_{1}=6, b_{2}=2, b_{3}=3$, and $b_{4}=-7$. Calculate the following sums. (Use symbolic notation and fractions where needed.) (a) $\sum_{i=2}^{4} b_{i}=$ (b) $\sum_{j=1}^{2}\left(b_{j}+2^{b_{j}}\right)=$ (c) $\sum_{k=1}^{3} k b_{k}=$

Step 1 ：Given the values of \(b_{i}\) where \(i = 1, 2, 3, 4\), we are asked to find the sum of \(b_{2}\), \(b_{3}\), and \(b_{4}\).

Step 2 ：Substituting the given values into the equation, we get \(\sum_{i=2}^{4} b_{i}= b_{2}+b_{3}+b_{4}=2+3+(-7)\).

Step 3 ：The sum is \(-2\). So, \(\sum_{i=2}^{4} b_{i} = \boxed{-2}\).

Step 4 ：Next, we are asked to find the sum of \(b_{1}+2^{b_{1}}\) and \(b_{2}+2^{b_{2}}\).

Step 5 ：Substituting the given values into the equation, we get \(\sum_{j=1}^{2}\left(b_{j}+2^{b_{j}}\right)= (b_{1}+2^{b_{1}})+(b_{2}+2^{b_{2}})=(6+2^{6})+(2+2^{2})\).

Step 6 ：The sum is \(76\). So, \(\sum_{j=1}^{2}\left(b_{j}+2^{b_{j}}\right) = \boxed{76}\).

Step 7 ：Finally, we are asked to find the sum of \(1*b_{1}\), \(2*b_{2}\), and \(3*b_{3}\).

Step 8 ：Substituting the given values into the equation, we get \(\sum_{k=1}^{3} k b_{k}= 1*b_{1}+2*b_{2}+3*b_{3}=1*6+2*2+3*3\).

Step 9 ：The sum is \(19\). So, \(\sum_{k=1}^{3} k b_{k} = \boxed{19}\).