What Is the Value of S_n in the Summation Formula?

[MarkFL]

Please compute the following sum:

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

 

[anemone]

Nice problem!:)

My solution:

Spoiler

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

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

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

Since \(\displaystyle {n\choose k}={n\choose n-k}\)

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

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

\(\displaystyle \;\;\;\;\;\;=\sum_{k=0}^{n} n {n\choose n-k}-\sum_{k=0}^{n} k {n\choose n-k}\)

\(\displaystyle \;\;\;\;\;\;=\sum_{k=0}^{n} n {n\choose k}-\sum_{k=0}^{n} k {n\choose k}\)

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

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

THus,

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

 

[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

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

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

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

 

[CellCoree]

Hello,

Thank you for your inquiry. To compute the sum S_n, we first need to understand the notation used.

The notation \sum_{k=1}^{n} represents the summation of a series, where k is the index and n is the number of terms in the series. In this case, the series starts at k=1 and ends at k=n.

The expression \frac{n!}{(k-1)!(n-k)!} represents the combination of n objects taken k at a time, also known as "n choose k". This can be calculated using the formula \binom{n}{k}=\frac{n!}{k!(n-k)!}.

Now, to compute S_n, we need to plug in the values for k in the given expression and sum them up. This can be done as follows:

S_n=\sum_{k=1}^{n}\frac{n!}{(k-1)!(n-k)!}=\binom{n}{1}+\binom{n}{2}+\binom{n}{3}+...+\binom{n}{n}

Using the formula for combinations, we can simplify this expression to:

S_n=n+n\cdot\frac{n-1}{2}+n\cdot\frac{n-1}{2}\cdot\frac{n-2}{3}+...+n\cdot\frac{n-1}{2}\cdot\frac{n-2}{3}\cdot...\cdot\frac{1}{n}

This simplification can be further written as:

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

Now, we can use the fact that n!=n\cdot(n-1)\cdot(n-2)\cdot...\cdot1 to rewrite the expression as:

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

Finally, we can use the identity \sum_{k=1}^{n}\frac{(n-1)!}{(n-k)!k!}=2^{n-1} to get:

S_n=n\cdot2^{n-1}

Therefore, the final expression for S_n is:

S_n=n\cdot2^{n-1}

I hope this helps in computing the given sum. Let me know if you have any further questions.

Best,
[Your