Solve : $\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$ Solve : $$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$$
Can we take like this : 
Let us take (k+1)th term  = $$\frac{(k+1)^2(k+1)!}{2^{k+1}} ; \frac{(k+1)(k+1)!}{2^{k+1}}$$ and (k-1)th  term is : $$\frac{(k-1)^2(k-1)!}{2^{k-1}};  \frac{(k-1)(k-1)!}{2^{k-1}}$$
What can we do further to this series... please suggest ......thanks.

Possible way:
$$\sum^{2010}_{k=1} \frac{k^2k!}{2^k}+\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}=\sum^{2010}_{k=1} \frac{k!(k+1)k}{2^k}=\sum^{2010}_{k=1} \frac{k(k+1)!}{2^k}$$
$$=\sum^{2010}_{k=1} \frac{(k+2-2)(k+1)!}{2^k}=\sum^{2010}_{k=1} \frac{(k+2)!}{2^k}-\sum^{2010}_{k=1} \frac{(k+1)!}{2^{k-1}}$$
$$=\frac{2012!}{2^{2010}}-2$$
