What is the value of $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~… ~ {}^nP_n/n $
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What is the value of $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
${}^nP_r = frac {n!} {(n-r)!}$
Attempt: $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
$= 1 + frac {n! }{2.(n-2)!} + frac {n! }{3.(n-3)!} + ........ + frac {n! }{n.(n-n)!}$
$ = 1 + {}^nC_2 + {}^nC_3 . 2! +~~ ............~~+ {}^nC_n. (n-1)! $ .........(1)
Now, $(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2x^2 + ............ + {}^nC_nx^n$
Diferentiating the above doesnt seem to work because in (1), every term has coffecient of the form ${}^nC_r r!$
Help shall be appreciated. Thanks
combinatorics permutations binomial-coefficients
$endgroup$
|
show 3 more comments
$begingroup$
What is the value of $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
${}^nP_r = frac {n!} {(n-r)!}$
Attempt: $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
$= 1 + frac {n! }{2.(n-2)!} + frac {n! }{3.(n-3)!} + ........ + frac {n! }{n.(n-n)!}$
$ = 1 + {}^nC_2 + {}^nC_3 . 2! +~~ ............~~+ {}^nC_n. (n-1)! $ .........(1)
Now, $(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2x^2 + ............ + {}^nC_nx^n$
Diferentiating the above doesnt seem to work because in (1), every term has coffecient of the form ${}^nC_r r!$
Help shall be appreciated. Thanks
combinatorics permutations binomial-coefficients
$endgroup$
$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as{}^nP_3
.
$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18
|
show 3 more comments
$begingroup$
What is the value of $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
${}^nP_r = frac {n!} {(n-r)!}$
Attempt: $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
$= 1 + frac {n! }{2.(n-2)!} + frac {n! }{3.(n-3)!} + ........ + frac {n! }{n.(n-n)!}$
$ = 1 + {}^nC_2 + {}^nC_3 . 2! +~~ ............~~+ {}^nC_n. (n-1)! $ .........(1)
Now, $(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2x^2 + ............ + {}^nC_nx^n$
Diferentiating the above doesnt seem to work because in (1), every term has coffecient of the form ${}^nC_r r!$
Help shall be appreciated. Thanks
combinatorics permutations binomial-coefficients
$endgroup$
What is the value of $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
${}^nP_r = frac {n!} {(n-r)!}$
Attempt: $1 + {{}^nP_2}/2 +{{}^nP_3}/3 + ~........... ~ {}^nP_n/n $
$= 1 + frac {n! }{2.(n-2)!} + frac {n! }{3.(n-3)!} + ........ + frac {n! }{n.(n-n)!}$
$ = 1 + {}^nC_2 + {}^nC_3 . 2! +~~ ............~~+ {}^nC_n. (n-1)! $ .........(1)
Now, $(1+x)^n = {}^nC_0 + {}^nC_1 x + {}^nC_2x^2 + ............ + {}^nC_nx^n$
Diferentiating the above doesnt seem to work because in (1), every term has coffecient of the form ${}^nC_r r!$
Help shall be appreciated. Thanks
combinatorics permutations binomial-coefficients
combinatorics permutations binomial-coefficients
edited Dec 22 '18 at 2:56
Shaun
9,804113684
9,804113684
asked Mar 28 '14 at 7:36
MathManMathMan
3,66042072
3,66042072
$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as{}^nP_3
.
$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18
|
show 3 more comments
$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as{}^nP_3
.
$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18
$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as
{}^nP_3
.$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as
{}^nP_3
.$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18
|
show 3 more comments
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$begingroup$
Have you calculated the first few values by hand to see (1) what sort of function you're probably looking for and (2) how those fractions will be made to have common denominators?
$endgroup$
– Eric Towers
Mar 28 '14 at 7:45
$begingroup$
What are $P_n2$? This looks like work for generating functions. Imagine a $x^k$ ($k=2,3,4...n$) next to every term, then differentiate, and then add by binomial formula.
$endgroup$
– orion
Mar 28 '14 at 7:45
$begingroup$
@orion $nP_r = frac {n!} {(n-r)!}$ . uhmm, you mean first differentiate each term and look for similar terms as in the question? It doesn't really seem to work
$endgroup$
– MathMan
Mar 28 '14 at 7:47
$begingroup$
Oops, doesn't really add up to what I wanted. Interesting question.
$endgroup$
– orion
Mar 28 '14 at 8:12
$begingroup$
This is extremely hard to read. I gather that in (for instance) $nP_3/3$ the $n$ is not a multiplicative factor but part of the symbol $nP_3$, which stands for $n(n-1)(n-2)=3!binom n3$. If you meant to write ${}^nP_3$ (which still is a horrible notation) instead of $nP_3$, type it as
{}^nP_3
.$endgroup$
– Marc van Leeuwen
Mar 28 '14 at 8:18