Combo Algebra

We have $(x+1)^{100}=\sum_{k=0}^{100}{100 \choose k}x^k$ . Differentiation gives $100(x+1)^{99}=\sum_{k=0}^{100}k{100 \choose k}x^{k-1}$ . Plug in $x=1$ to find $\sum_{k=0}^{100}k{100 \choose k}=100(2^{99})=25(2^{101})$ , with $25+101=\boxed{126}$

We will count the expression in two ways. Notice that if we have a committee of $100$ people, the given expression is just the number of ways to choose a non-empty subcommittee, and then choose a president amoung the subcommittee.

However, this is equivalent to choosing the president first, and then deciding the people who are going to be in the subcommitte.

We can choose the president in $100$ ways. Each of the other people have two choices: they are either in the subcommittee or they are not in the subcommitte, therefore this is $2^{99}$ for all 99 other people.

Therefore, the number of ways is just $100 \cdot 2^{99} = 25 \cdot 2^{101} \implies 25 + 101 = \boxed{126}$

Note this gives us an identity: $\sum_{k = 1}^{n}{k{n \choose k}} = n \cdot 2^{n-1}$

by the binomial theorm: $\sum_{n=0}^{100} \binom{100}{n} = 2^{100}$ we need to find $S=\sum_{n=0}^{100} n\binom{100}{n}$ this equals $S=0\binom{100}{0}+1\binom{100}{1}+2\binom{100}{2}+....+98\binom{100}{98}+99\binom{100}{99}+100\binom{100}{100}$ take all n to be 100: $S=100\binom{100}{0}+100\binom{100}{1}+....+100\binom{100}{99}+100\binom{100}{100}-(100\binom{100}{0}+99\binom{100}{1}+...+1\binom{100}{99}+0\binom{100}{100})$ since $\binom{n}{k}=\binom{n}{n-k}$ : $S=100*2^{100}-(100\binom{100}{100}+99\binom{100}{99}+...+1\binom{100}{1}+0\binom{100}{0}$ $S=100*2^{100}-S$ $S=100*2^{99}=25*2^{101}$ $25+101=\boxed{126}$

Note that $\dbinom{n}{0} + \dbinom{n}{1} + \dbinom{n}{2} + ... + \dbinom{n}{n} = 2^n$ . From this fact, we do some algebra: $\dbinom{100}{1} + 2\dbinom{100}{2} + ... + 100\dbinom{100}{100} = \displaystyle\sum\limits_{k=1}^{100} k\dbinom{100}{k} = \sum\limits_{k=1}^{100} k\frac{100!}{(100-k)!k!}$ $= 100 \sum\limits_{k=1}^n \frac{(100-1)!}{(100-k)!(k-1)!} = 100 \sum\limits_{k=1}^n \dbinom{99}{k-1} = 100 \cdot 2^{99} = 25 \cdot 2^{101}.$ So our final answer must be $25 + 101 = \boxed{126}$ .

Take note that $\left( \begin{matrix} n \\ r \end{matrix} \right) =\left( \begin{matrix} n \\ n-r \end{matrix} \right)$

Rewrite the equation, we get $\left( \begin{matrix} 100 \\ 99 \end{matrix} \right) +2\left( \begin{matrix} 100 \\ 98 \end{matrix} \right) +3\left( \begin{matrix} 100 \\ 97 \end{matrix} \right) +...+100\left( \begin{matrix} 100 \\ 0 \end{matrix} \right)$

By rearranging, we get $100(\left( \begin{matrix} 100 \\ 0 \end{matrix} \right) +99\left( \begin{matrix} 100 \\ 1 \end{matrix} \right) +98\left( \begin{matrix} 100 \\ 2 \end{matrix} \right) +...+\left( \begin{matrix} 100 \\ 99 \end{matrix} \right)$

By adding with the original equation, we get $100(\left( \begin{matrix} 100 \\ 0 \end{matrix} \right) +100\left( \begin{matrix} 100 \\ 1 \end{matrix} \right) +100\left( \begin{matrix} 100 \\ 2 \end{matrix} \right) +...+100\left( \begin{matrix} 100 \\ 99 \end{matrix} \right) +100\left( \begin{matrix} 100 \\ 100 \end{matrix} \right) =100\sum _{ r=0 }^{ 100 }{ \left( \begin{matrix} 100 \\ r \end{matrix} \right) }$

It is known that $\sum _{ r=0 }^{ n }{ \left( \begin{matrix} n \\ r \end{matrix} \right) } ={ 2 }^{ n }$

Thus, we get $2k\cdot2^{n}=100\cdot2^{100}$

$k\cdot2^n=50\cdot2^{100}=25\cdot2^{101}$

Thus, $k+n=\boxed{126}$

$\begin{aligned} (1+x)^{100} & = \begin{pmatrix} 100 \\ 0 \end{pmatrix} + \begin{pmatrix} 100 \\ 1 \end{pmatrix} x + \begin{pmatrix} 100 \\ 2 \end{pmatrix} x^2 +... + \begin{pmatrix} 100 \\ 100 \end{pmatrix}x^{100} \\ \Rightarrow \frac{d}{dx} (1+x)^{100} & = \begin{pmatrix} 100 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} 100 \\ 2 \end{pmatrix} x + 3 \begin{pmatrix} 100 \\ 3 \end{pmatrix} x^2 +... + 100 \begin{pmatrix} 100 \\ 100 \end{pmatrix}x^{99} \quad \quad \color{#3D99F6}{\text{Put } x = 1} \\ 100 (1+\color{#3D99F6}{[1]})^{99} & = \begin{pmatrix} 100 \\ 1 \end{pmatrix} + 2 \begin{pmatrix} 100 \\ 2 \end{pmatrix}\color{#3D99F6}{[1]} + 3 \begin{pmatrix} 100 \\ 3 \end{pmatrix} \color{#3D99F6}{[1]} +... + 100 \begin{pmatrix} 100 \\ 100 \end{pmatrix}\color{#3D99F6}{[1]} \\ & = 100\dot{}2^{99} = 25\dot{}2^{101} \\ & \\ \Rightarrow k + n & = 25 + 101 = \boxed{126} \end{aligned}$

We have $\dbinom{n}{m} = \dbinom{n}{m - n}$ and $\dbinom{n}{1} + \dbinom{n}{2} + \cdots + \dbinom{n}{n-1} + \dbinom{n}{n} = 2^{n}$

Thus $X = \dbinom{100}{1} + 2\dbinom{100}{2} + \cdots + 100\dbinom{100}{100} = 100\Bigg( \dbinom{100}{0} + \dbinom{100}{1} + \cdots + \dbinom{100}{49}\Bigg) + 50 \dbinom{100}{50}$

$X = 50\Bigg( 2\bigg(\dbinom{100}{0} + \dbinom{100}{1} + \cdots + \dbinom{100}{49}\bigg) + \dbinom{100}{50}\Bigg)$

$X = 50 \cdot 2^{100}$

$X = 25 \cdot 2^{101}$