A Sum involving Binomial Coefficients

Consider: $(x+1)^n= \sum_{r=0}^{r=n} x^r\binom{n}{r}$

Differentiate both sides w.r.t $x$

$\implies n(x+1)^{n-1}= \sum_{r=0}^{r=n} rx^{r-1}\binom{n}{r}$

Put $x=1$ to get:

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

$\therefore f(n)=n2^{n-1}\implies f(100)=100\cdot 2^{99}=25\cdot 2^{\color{#D61F06}{101}}$

Therefore the largest integer $k$ such that $2^k \mid f(100)$ is $\boxed{\color{#D61F06}{101}}$ .

My solution:

We know that ${n \choose r} = {n \choose n-r}$ , which implies that

$\large \displaystyle f(n)= \sum_{r=0}^{n} r\binom{n}{r} = \sum_{r=0}^{n} r\binom{n}{n-r}$

Which in turn implies that

$\large \displaystyle 2f(n) = \sum_{r=0}^{n} r\binom{n}{r} + \color{#3D99F6}{ \sum_{r=0}^{n} r\binom{n}{n-r}}$ $\large \displaystyle = \sum_{r=0}^{n} r\binom{n}{r} + \color{#3D99F6}{ \sum_{r=0}^{n} (n-r) \binom{n}{r}}$ $= \large \displaystyle n \sum_{r=0}^{n} {n \choose r} = n(2^n)$

And so, $f(n) = n(2^{n-1})$ .

We are then looking for the largest power of two which divides $f(100) = 100(2^{99}) = 25(2^{101})$ . We can then easily see that this power is $2^{101}$ , so our final answer is $\boxed{101}$ .

@Sambhrant Sachan and @Rishabh Cool have already shown solutions with the help of Calculus.

Initially I had another solution. Let me demonstrate it for even $n$ . For odd $n$ , it's pretty similar.

We'll use this Combinatorial Identity: $n\binom{n}{r}=r\binom{n}{r}+(r+1)\binom{n}{r+1}\cdots{ } \cdots{ } \cdots{ }(1)$

We have, $f(n)$

$= \sum_{r=0}^{n} r\binom{n}{r}$

$= \sum_{r=1}^{n} r\binom{n}{r}$

$= \left(1\binom{n}{1}+2\binom{n}{2}\right)+\left(3\binom{n}{3}+4\binom{n}{4}\right)+\cdots{ } \cdots{ } \cdots{ }+\left((n-1)\binom{n}{n-1}+n\binom{n}{n}\right)$

$= n\binom{n}{1}+n\binom{n}{3}+\cdots{ } \cdots{ } \cdots{ }+n\binom{n}{n-1}$

$=n \left( \binom{n}{1}+\binom{n}{3}+\cdots{ } \cdots{ } \cdots{ }+\binom{n}{n-1}\right)$

$=n\times 2^{n-1}$ .

Then $f(100)=100\times 2^{99}=25\times 2^{101}$ .

So, $\boxed{101}$ is the answer.

Later when I tried to generalize it for $\sum_{r=0}^{n} r^k\binom{n}{r}$ , I found the Calculus approach to be better.

starting with binomial expansion of $(1+x)^{n}$ . Where $x$ is a complex number and $n$ is a whole number.

$\ (1+x)^{n} = \dbinom{n}{0}x^{0}+\dbinom{n}{1}x^{1}+\dbinom{n}{2}x^{2}+\cdots+\dbinom{n}{n-1}x^{n-1}+\dbinom{n}{n}x^{n} = \displaystyle\sum_{r=0}^{n}\dbinom{n}{r}x^{r}$

Differentiate the equation with respect to $x$

$\ n(1+x)^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}x^{1}+\cdots+(n-1)\cdot\dbinom{n}{n-1}x^{n-2}+n\cdot\dbinom{n}{n}x^{n-1} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r}x^{r-1}$

put $x=1$

$\boxed{ n\cdot2^{n-1} =1\cdot\dbinom{n}{1}+2\cdot\dbinom{n}{2}+\cdots+(n-1)\cdot\dbinom{n}{n-1}+n\cdot\dbinom{n}{n} = \displaystyle\sum_{r=0}^{n}r\dbinom{n}{r} }$

$f(100) = 100\times 2^{99} = 25 \times 2^{101}$

Therefore the largest integer $k$ must be equal to $101$

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Check out my note on Laws of binomial theorem