10 find ∑ (k+2)*3*2^k k=0

           10
find      ∑ (k+2)*3*2^k
         k=0

$\begin{array}{|rcll|} \hline && \sum \limits_{k=0}^{10} {(k+2)\cdot 3 \cdot 2^k} \\ &=& 3\cdot \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} \\ &=& 3\cdot S_{10} \qquad |\qquad \text{We set } ~ S_{10} = \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} \\ \hline \end{array}$

$\begin{array}{|lcll|} \hline S_{10} = \sum \limits_{k=0}^{10} {(k+2)\cdot 2^k} &=& 2\cdot 2^0 + & 3\cdot 2^1+ 4\cdot 2^2+ \dots + 11\cdot 2^9 + 12\cdot 2^{10} \\ 2\cdot S_{10} &=& & 2\cdot 2^1 + 3\cdot 2^2 + \dots + 10\cdot 2^{9}+ 11\cdot 2^{10} + 12\cdot 2^{11} \\ \hline S_{10} -2\cdot S_{10} &=& 2\cdot 2^0 & +2^1+2^2+2^3+\dots + 2^9+2^{10}-12\cdot2^{11} \\\\ && & 2^0+2^1+2^2+2^3+\dots + 2^9+2^{10} = 2^{11} - 1 \\ && & 2^1+2^2+2^3+\dots + 2^9+2^{10} = 2^{11} - 2 \\\\ S_{10} -2\cdot S_{10} &=& 2\cdot 2^0 & +2^{11} - 2 -12\cdot2^{11} \\ (1-2)\cdot S_{10} &=& 2\cdot 2^0 & +2^{11} - 2 -12\cdot2^{11} \\ - S_{10} &=& & 2^{11} -12\cdot2^{11} \\ - S_{10} &=& & -2^{11}\cdot ( 12-1) \\ S_{10} &=& & 2^{11}\cdot 11 \\ S_{10} &=& & 11\cdot 2^{11} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline && \sum \limits_{k=0}^{10} {(k+2)\cdot 3 \cdot 2^k} \\ &=& 3\cdot S_{10} \\ &=& 3\cdot 11\cdot 2^{11} \\ &=& 33\cdot 2^{11} \\ &=& 67584 \\ \hline \end{array}$

sum_(k=0)^10 3 (k+2) 2^k = 67,584

           10
find      ∑ (k+2)*3*2^k
         k=0

$\displaystyle\sum_{k=0}^{10}\;\;3(k+2)*2^k\;\;\\ =3*\displaystyle\sum_{k=0}^{10}\;\;(k+2)*2^k\;\;\\ =3[(2)+(3*2)+(4*4)+(5*8)+(6*16)+(7*32)+(8*64)+(9*128)+(10*256)+(11*512)+(12*1024)]$

=67584