A Twisted Binomial Expansion!

Let $S = f(a,b,n)$ . Then,

$\large{S = \sum_{r=0}^n {n \choose r} \log \left(a^{n-r} b^r \right) = \log \left( \prod_{r=0}^n \left( a^{n-r} b^r \right)^{{n \choose r}} \right)}$

$\large{=\log \left( a^{\sum (n-r) {n \choose r}} b^{\sum r {n \choose r}} \right) \quad ...(1)}$

where both summations in the above line are from $r=1$ to $r=n$ .

Using the standard well known method, we differentiate the binomial expansion: $(1+x)^n = \sum_{r=0}^n {n \choose r} x^r$ to obtain:

$n(1+x)^{n-1} = \sum_{r=1}^n r{n \choose r} x^{r-1}$

Multiplying the above equation by $x$ , we have $nx(1+x)^{n-1} = \sum_{r=1}^n r{n \choose r} x^r$

Setting $x=1$ , we then obtain:

$\sum_{r=0}^n r {n \choose r} = \sum_{r=1}^n r {n \choose r} = n2^{n-1}$

Hence,

$\sum_{r=0}^n (n-r) {n \choose r} = n\sum_{r=0}^n {n \choose r} - \sum_{r=0}^n r {n \choose r}$

$= n2^n - n2^{n-1} = n2^{n-1}$

Substituting them in equation $(1)$ , we obtain $S = n2^{n-1} \cdot \log(ab)$ .

Putting $a=7, b=11, n=21$ , we obtain the expression as: $\large{2^{20} \cdot 3^1 \cdot 7^1 \cdot \log(77)}$ , $\quad$ and thus:

$2+3+7+20+1+1+77 = \boxed{111}$

one can do it by replacing r by n-r and adding new one with old ones then we get 2S=(2^n) (log[(ab)^n]) . NOTE - * 2^n is summation of nCr where r is from 0 to n *

IF original then great job