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- Thread starter MarkFL
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- Feb 14, 2012

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Nice problem!

My solution:

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,

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

My solution:

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,

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

Last edited:

- Mar 31, 2013

- 1,347

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

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Here is my solution:

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

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