What Is the Value of S_n in the Summation Formula?

[MarkFL]

Please compute the following sum:

$$S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}$$

 

[anemone]

Nice problem!:)

My solution:

Spoiler

We're given $$S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}$$.

By multiplying the variable $k$ on top and bottom of the fraction, we get

$$\small S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}=\sum_{k=1}^{n}\frac{k(n!)}{k(k-1)!(n-k)!}=\sum_{k=1}^{n}\frac{k(n!)}{(k)!(n-k)!}=\sum_{k=1}^{n} k {n\choose k}=\sum_{k=0}^{n} k {n\choose k}-0{n\choose k}=\sum_{k=0}^{n} k {n\choose k}$$

Since $${n\choose k}={n\choose n-k}$$

We see that there is another way to rewrite $S_n$, i.e.

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

$$\;\;\;\;\;\;=\sum_{k=0}^{n} n {n\choose n-k}-\sum_{k=0}^{n} k {n\choose n-k}$$

$$\;\;\;\;\;\;=\sum_{k=0}^{n} n {n\choose k}-\sum_{k=0}^{n} k {n\choose k}$$

$$\;\;\;\;\;\;=\sum_{k=0}^{n} n {n\choose k}-S_n$$

$$\therefore 2S_n=\sum_{k=0}^{n} n {n\choose k}=n\sum_{k=0}^{n} {n\choose k}=n(2^n)$$

THus,

$$\therefore S_n=n(2)^{n-1}$$

 

[kaliprasad]

Good ans by anemone .

Here is mine
anemone has shown that

Sn = ( k = 1 to n) ∑ k(nCk)

We know

(x+1)^n = ( k = 0 to n) ∑ (nCk)x^k

Differentiate both sides wrt x

n(x+1)^(n-1) = ( k = 1 to n) ∑ k (nCk)x^(k-1) knowing that d/dx(x^0) = 0 so it is dropped

put x = 1 on both sides to get

n 2^(n-1) = ( k = 1 to n) ∑ k (nCk) =Sn

 

[MarkFL]

Thank you anemone and kaliprasad for participating! (Sun)

Here is my solution:

Spoiler

$$S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}$$

$$S_n=\sum_{k=0}^{n-1}\frac{n!}{((k+1)-1)!(n-(k+1))!}=n\sum_{k=0}^{n-1}\frac{(n-1)!}{k!((n-1)-k)!}$$

$$S_n=n\sum_{k=0}^{n-1}{n-1 \choose k}=n(1+1)^{n-1}=n2^{n-1}$$