If $f(n) =\displaystyle\sum_{r=1}^{n}\Biggl(r^n\Bigg(\binom{n}{r}-\binom{n}{r-1}\Bigg) + (2r+1)\binom{n}{r}\Biggr)$, then what is $f(30)$?

Question

Please give me hints on how to solve it. I tried 2-3 methods but it doesn't go beyond two steps. I am out of ideas now.

Thank you

Answer 1

We may simply deal with each piece separately: $$ \sum_{r=1}^{n}(2r+1)\binom{n}{r}=\left.\frac{d}{dx}\sum_{r=1}^{n}\binom{n}{r}x^{2r+1}\right|_{x=1}=\left.\frac{d}{dx}\left(x\cdot\left(1+x^2\right)^n-1\right)\right|_{x=1}=2^n(n+1)-1.$$ On the other hand, by summation by parts: $$ T_n=\sum_{r=1}^{n}\left(\binom{n}{r}-\binom{n}{r-1}\right)r^n \\= n^n\left(\binom{n}{n}-\binom{n}{0}\right)-\sum_{r=1}^{n-1}\left(\binom{n}{r}-1\right)((r+1)^n-r^n) $$ so: $$ T_n = n^n-1-\sum_{r=1}^{n-1}\binom{n}{r}((r+1)^n-r^n) $$ and you may compute the last sum through Dobinski's formula.