Solving equation of summation of n combinations

Exercise:

Solve the following equation $\sum\limits_{k=1}^{n}{\left( n-k \right)} {{n}\choose{k}}=441$

Solution: we have $\sum\limits_{k=1}^{n}{\left( n-k \right)} {{n}\choose{k}}=441\Leftrightarrow \sum\limits_{k=1}^{n} n {{n}\choose{k}}-\sum\limits_{k=1}^{n} k {{n}\choose{k}}=441$

We know that from Binomial theorem ${(x+y)}^{n}=\sum\limits_{k=0}^{n} {{n}\choose{k}}{x}^{n-k}{y}^{k}$ where $k\leq n$

Put $x=1 \,\, \&\,\,   y=1 $ to get $\sum\limits_{k=0}^{n}{{n}\choose{k}}=2^n \Leftrightarrow {{n}\choose{0}}+\sum\limits_{k=1}^{n}=2^n$ where ${{n}\choose{0}}=1$

So $\sum\limits_{k=1}^{n}{{n}\choose{k}}=2^n-1$ thus $\sum\limits_{k=1}^{n}n{{n}\choose{k}}=n\sum\limits_{k=1}^{n}{{n}\choose{k}}= n(2^n-1)=n2^n-n   .... (*)$

Notice that, ${{n}\choose{k}}=\frac{n!}{(n-k)!k!}=\frac{n(n-1)!}{k(k-1)!(((n-1)-(k-1))!}=\frac{n}{k}{{n-1}\choose{k-1}}$

hence $k{{n}\choose{k}}=n{{n-1}\choose{k-1}}\Leftrightarrow \sum\limits_{k=1}^{n}k{{n}\choose{k}} = \sum\limits_{k=0}^{n-1}n{{n-1}\choose{k-1}}=n2^{n-1}  ....(**)$

Thus $\sum\limits_{k=1}^{n}n{{n}\choose{k}}-\sum\limits_{k=1}^{n}k{{n}\choose{k}}=n2^n-n-n2^{n-1}=441$

$\Rightarrow n{{2}^{n}}-n-\frac{n{{2}^{n}}}{2}=441\Rightarrow {{2}^{n}}\left( n-\frac{n}{2} \right)=441+n\Leftrightarrow {{2}^{n}}\left( \frac{n}{2} \right)=441+n$

$\Rightarrow n{{2}^{n}}=882+2n\Rightarrow n\left( {{2}^{n}}-2 \right)=882$

By using numerical method we get $n=7$