$${{a}_{n}}={{2}^{n}}$$，$${{b}_{n}}=n$$，$${{A}_{n}}={{2}^{n+1}}-2$$，$${{B}_{n}}=\frac{n\left( n+1 \right)}{2}$$，于是$${{c}_{n}}=\frac{n\left( n+1 \right)}{2}{{2}^{n}}+n\cdot {{2}^{n}}-2n$$，  不妨设$${{P}_{n}}=\sum\limits_{i=1}^{n}{i\cdot {{2}^{i}}}$$，$${{Q}_{n}}=\sum\limits_{i=1}^{n}{\frac{i\left( i+1 \right)}{2}\cdot {{2}^{i}}}$$，于是$$\sum\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\left( n+1 \right)}$$，  $${{Q}_{n}}=1\times 2+3\times {{2}^{2}}+\cdots +\frac{n\left( n+1 \right)}{2}\cdot {{2}^{n}}$$①  $$2{{Q}_{n}}=1\times {{2}^{2}}+3\times {{2}^{3}}+\cdots +\frac{n\left( n-1 \right)}{2}\cdot {{2}^{n}}+\frac{n\left( n+1 \right)}{2}\cdot {{2}^{n+1}}$$②  ①$$-$$②得$$-{{Q}_{n}}=1\times {{2}^{1}}+2\times {{2}^{2}}+\cdots +n\cdot {{2}^{n}}-\frac{n\left( n+1 \right)}{2}\cdot {{2}^{n+1}}$$  即$${{Q}_{n}}=\frac{n\left( n+1 \right)}{2}\cdot {{2}^{n+1}}-{{P}_{n}}$$，即$${{Q}_{n}}+{{P}_{n}}=\frac{n\left( n+1 \right)}{2}\cdot {{2}^{n+1}}=n\left( n+1 \right)\cdot {{2}^{n}}$$  ∴$$\sum\limits_{i=1}^{n}{{{c}_{i}}={{P}_{n}}+{{Q}_{n}}-n\left( n+1 \right)=n\left( n+1 \right)\left( {{2}^{n}}-1 \right)}$$，将$$n=10$$代入即可．  故选$$\text{D}$$． 
设$${{a}_{n}}={{2}^{n}}$$，$${{b}_{n}}=n$$，（$$n=1$$，$$2$$，$$3$$\ldots），$${{A}_{n}}$$、$${{B}_{n}}$$分别为数列$$\left { {{a}_{n}} \right }$$、$$\left { {{b}_{n}} \right }$$的前$$n$$项和，记$${{c}_{n}}={{a}_{n}}{{B}_{n}}+{{b}_{n}}{{A}_{n}}-{{a}_{n}}{{b}_{n}}$$，则数列$$\left { {{c}_{n}} \right }$$的前$$10$$项和为（ ~ ~）．
[1] $${{2}^{10}}+53$$ 
[2] $${{2}^{11}}+53$$ 
[3] $$110\times \left( {{2}^{9}}-1 \right)$$ 
[4] $$110\times \left( {{2}^{10}}-1 \right)$$~ 