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Evaluate: $sum_k=2^n n!over (n-k)!(k-2)!$

2

$begingroup$

Evaluate:

$$sum_k=2^n n!over (n-k)!(k-2)!$$

Efforts:

To be honest, I just played with the expression, but I don't know how to approach it.

Take $n=2$,

we get $2!$

Take $n=3$, we get

$3!over 1!0!+ 3!over 0!1!$

Take $n=4$

we get $4!over 2!0!+ 4!over 1!1!+ 4!over 0!2!$

Take $n=5$, we get $5!over 3!0!+ 5!over 2!1!+5!over 1!2!+5!over 0!3!$

So this is what I have observed,

for each $n$, we have $n!$ in the denominator. And in denominator, we have terms like $a!b!$ such that $a+b=n-k$

Example in case of $N=6$ we have $6!$ in numerator. In denominator we have terms like $4!0!, 3!1!, 2!2!, 1!3!, 0!4!$

but how does that help. There is clearly a pattern in how the denominator appears but does that help in some way?

real-analysis

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asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

add a comment | 

2

$begingroup$

Evaluate:

$$sum_k=2^n n!over (n-k)!(k-2)!$$

Efforts:

To be honest, I just played with the expression, but I don't know how to approach it.

Take $n=2$,

we get $2!$

Take $n=3$, we get

$3!over 1!0!+ 3!over 0!1!$

Take $n=4$

we get $4!over 2!0!+ 4!over 1!1!+ 4!over 0!2!$

Take $n=5$, we get $5!over 3!0!+ 5!over 2!1!+5!over 1!2!+5!over 0!3!$

So this is what I have observed,

for each $n$, we have $n!$ in the denominator. And in denominator, we have terms like $a!b!$ such that $a+b=n-k$

Example in case of $N=6$ we have $6!$ in numerator. In denominator we have terms like $4!0!, 3!1!, 2!2!, 1!3!, 0!4!$

but how does that help. There is clearly a pattern in how the denominator appears but does that help in some way?

real-analysis

share|cite|improve this question

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

add a comment | 

$begingroup$

Evaluate:

$$sum_k=2^n n!over (n-k)!(k-2)!$$

Efforts:

To be honest, I just played with the expression, but I don't know how to approach it.

Take $n=2$,

we get $2!$

Take $n=3$, we get

$3!over 1!0!+ 3!over 0!1!$

Take $n=4$

we get $4!over 2!0!+ 4!over 1!1!+ 4!over 0!2!$

Take $n=5$, we get $5!over 3!0!+ 5!over 2!1!+5!over 1!2!+5!over 0!3!$

So this is what I have observed,

for each $n$, we have $n!$ in the denominator. And in denominator, we have terms like $a!b!$ such that $a+b=n-k$

Example in case of $N=6$ we have $6!$ in numerator. In denominator we have terms like $4!0!, 3!1!, 2!2!, 1!3!, 0!4!$

but how does that help. There is clearly a pattern in how the denominator appears but does that help in some way?

real-analysis

share|cite|improve this question

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

$endgroup$

share|cite|improve this question

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

share|cite|improve this question

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

asked Mar 27 at 7:56

StammeringMathematicianStammeringMathematician

2,8171324

2 Answers
2

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3

$begingroup$

$$sum_k=2^nfracn!left(n-kright)!left(k-2right)!=nleft(n-1right)sum_k=2^nbinomn-2k-2=nleft(n-1right)sum_k=0^n-2binomn-2k=nleft(n-1right)2^n-2$$

share|cite|improve this answer

answered Mar 27 at 8:01

drhabdrhab

104k545136

$endgroup$

add a comment | 

2

$begingroup$

You may procced as follows:

• $fracn!(n-k)!(k-2)!=binomnk cdot k(k-1)$


• $[(1+x)^n]'' = sum_k=2^nbinomnkcdot k(k-1)x^k-2$

It follows

$$n(n-1)2^n-2 =n(n-1)(1+1)^n-2 = sum_k=2^nbinomnkcdot k(k-1) = sum_k=2^n n!over (n-k)!(k-2)!$$

share|cite|improve this answer

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

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add a comment | 

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3

$begingroup$

$$sum_k=2^nfracn!left(n-kright)!left(k-2right)!=nleft(n-1right)sum_k=2^nbinomn-2k-2=nleft(n-1right)sum_k=0^n-2binomn-2k=nleft(n-1right)2^n-2$$

share|cite|improve this answer

answered Mar 27 at 8:01

drhabdrhab

104k545136

$endgroup$

add a comment | 

3

$begingroup$

$$sum_k=2^nfracn!left(n-kright)!left(k-2right)!=nleft(n-1right)sum_k=2^nbinomn-2k-2=nleft(n-1right)sum_k=0^n-2binomn-2k=nleft(n-1right)2^n-2$$

share|cite|improve this answer

answered Mar 27 at 8:01

drhabdrhab

104k545136

$endgroup$

add a comment | 

$begingroup$

$$sum_k=2^nfracn!left(n-kright)!left(k-2right)!=nleft(n-1right)sum_k=2^nbinomn-2k-2=nleft(n-1right)sum_k=0^n-2binomn-2k=nleft(n-1right)2^n-2$$

share|cite|improve this answer

answered Mar 27 at 8:01

drhabdrhab

104k545136

$endgroup$

share|cite|improve this answer

answered Mar 27 at 8:01

drhabdrhab

104k545136

answered Mar 27 at 8:01

drhabdrhab

104k545136

answered Mar 27 at 8:01

drhabdrhab

104k545136

2

$begingroup$

You may procced as follows:

• $fracn!(n-k)!(k-2)!=binomnk cdot k(k-1)$


• $[(1+x)^n]'' = sum_k=2^nbinomnkcdot k(k-1)x^k-2$

It follows

$$n(n-1)2^n-2 =n(n-1)(1+1)^n-2 = sum_k=2^nbinomnkcdot k(k-1) = sum_k=2^n n!over (n-k)!(k-2)!$$

share|cite|improve this answer

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

$endgroup$

add a comment | 

2

$begingroup$

You may procced as follows:

• $fracn!(n-k)!(k-2)!=binomnk cdot k(k-1)$


• $[(1+x)^n]'' = sum_k=2^nbinomnkcdot k(k-1)x^k-2$

It follows

$$n(n-1)2^n-2 =n(n-1)(1+1)^n-2 = sum_k=2^nbinomnkcdot k(k-1) = sum_k=2^n n!over (n-k)!(k-2)!$$

share|cite|improve this answer

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

$endgroup$

add a comment | 

$begingroup$

You may procced as follows:

• $fracn!(n-k)!(k-2)!=binomnk cdot k(k-1)$


• $[(1+x)^n]'' = sum_k=2^nbinomnkcdot k(k-1)x^k-2$

It follows

$$n(n-1)2^n-2 =n(n-1)(1+1)^n-2 = sum_k=2^nbinomnkcdot k(k-1) = sum_k=2^n n!over (n-k)!(k-2)!$$

share|cite|improve this answer

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

$endgroup$

share|cite|improve this answer

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929

answered Mar 27 at 8:10

trancelocationtrancelocation

14.3k1929