Answer
160
Work Step by Step
$\sum ^{6}_{k=1}\left( k+1\right) 2^{k-1}=\left( 1+1\right) \times 2^{1-1}+\left( 2+1\right) \times 2^{2-1}+\left( 3+1\right) \times 2^{3-1}+\left( 4+1\right) \times 2^{4-1}+\left( 5+1\right) \times 2^{5-1}+\left( 6+1\right) \times 2^{6-1} =2\times 2^{\circ }+3\times 2^{1}+4\times 2^{2}+5\times 2^{3}+6\times 2^{4}=2+6+16+40+96=160 $
