A Question on Polynomials, part 3

To construct such a polynomial $f$ , we have to note that $\displaystyle \sum_{r=0}^{n}{n \choose r} =2^{n}$ and that $\displaystyle \sum_{r=0}^{n}r{n \choose r} =n2^{n-1}$ .

Now, consider $\displaystyle f(x)=4\sum_{r=0}^{100}r{x-1 \choose r} +2\sum_{r=0}^{100}{x-1 \choose r}$ . For each integer $1\le k \le 101$ , $f(k)=4\sum_{r=0}^{100}r{k-1 \choose r} +2\sum_{r=0}^{100}{k-1 \choose r}$ $= 4 (k-1) 2^{k-2} + 2\times 2^{k-1}$ $= (k-1)2^{k} +2^{k} =k\times 2^{k}$

So, $f(102)=4\sum_{r=0}^{100}r{101 \choose r} +2\sum_{r=0}^{100}{101 \choose r}$ $=4\sum_{r=0}^{101}r{101 \choose r} +2\sum_{r=0}^{101}{101 \choose r} - 4(101)-2(1)$ $= 102 \times 2^{102} - 406$