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.
 A: Possible way:
$$\sum^{2010}_{k=1} \frac{k^2k!}{2^k}+\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}=\sum^{2010}_{k=1} \frac{k!(k+1)k}{2^k}=\sum^{2010}_{k=1} \frac{k(k+1)!}{2^k}$$
$$=\sum^{2010}_{k=1} \frac{(k+2-2)(k+1)!}{2^k}=\sum^{2010}_{k=1} \frac{(k+2)!}{2^k}-\sum^{2010}_{k=1} \frac{(k+1)!}{2^{k-1}}$$
$$=\frac{2012!}{2^{2010}}-2$$
A: The "2010" in
$\frac{2012!}{2^{2010}}-\sum^{2010}_{k=1} \frac{k^2k!}{2^k}-\sum^{2010}_{k=1} \frac{k\cdot k!}{2^k}$
is a MacGuffin.
If it is replaced by $n$,
this becomes
$
\frac{(n+2)!}{2^n}-\sum^{n}_{k=1} \frac{k^2k!}{2^k}-\sum^{n}_{k=1} \frac{k\cdot k!}{2^k}
$.
Adding the last two terms
(as Kunnysan did)
$\begin{align}
\sum^{n}_{k=1} \frac{k^2k!}{2^k}+\sum^{n}_{k=1} \frac{k\cdot k!}{2^k}
&=\sum^{n}_{k=1} \frac{k^2k!+k\cdot k!}{2^k}\\
&=\sum^{n}_{k=1} k!\frac{k^2+k}{2^k}\\
&=\sum^{n}_{k=1} k!\frac{k(k+1)}{2^k}\\
&=\sum^{n}_{k=1} (k+1)!\frac{k}{2^k}\\
&=\sum^{n}_{k=1} (k+1)!\frac{(k+2)-2}{2^k}
\quad \text{This is Kunnysan's very ingenious key step}\\
&=\sum^{n}_{k=1} \frac{(k+2)!}{2^k}
-\sum^{n}_{k=1} \frac{(k+1)!}{2^{k-1}}\\
&=\sum^{n+1}_{k=2} \frac{(k+1)!}{2^{k-1}}
-\sum^{n}_{k=1} \frac{(k+1)!}{2^{k-1}}\\
&=\frac{(n+2)!}{2^{n}}
-\frac{(2)!}{2^{0}}\\
&=\frac{(n+2)!}{2^{n}}
-2\\
\end{align}
$
so the result is $2$
independent of $n$.
