Why $sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = 1/(1-k)$? The Next CEO of Stack Overflowmaybe this sum have approximation $sum_k=0^nbinomnk^3approxfrac2pisqrt3ncdot 8^n,ntoinfty$How prove binomial cofficients $sum_k=0^[fracn3](-1)^kbinomn+1kbinom2n-3kn=sum_k=[fracn2]^nbinomn+1kbinomkn-k$How find the sum $2sum_k=1^inftysum_i=0^2k-1fracbinom2kicdot B_icdot(m-1)^2k-i2k(2k-1)m^2k-1$How to prove that $sum_k=0^infty binomxx-kcdotbinomxk-x = 1$?prove that $(frac16)^4cdotlim_nrightarrowinftysum_i=4^nbinomi-13(frac56)^i-4=1$Evaluate$sum_n=0^inftybinom3nnx^n$$sum_i=1^ni^k=sum_i=1^kfracc_k+1-ik!i!n^i$. What is known about $c_k_k=0^infty$?$lim_limitssigma_A, sigma_B to infty e^-sigma_A-sigma_B sum_k=0^infty fracsigma_A^kk!cdotfracsigma_B^k+1(k+1)!$Proof for binomial theorem $ (1+x)^a = sum_k=0^infty P_k(a)x^k $Help understanding proof of the following statement $E(Y) = sum_i = 1^infty P(Y geq k)$
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Why $sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = 1/(1-k)$?
The Next CEO of Stack Overflowmaybe this sum have approximation $sum_k=0^nbinomnk^3approxfrac2pisqrt3ncdot 8^n,ntoinfty$How prove binomial cofficients $sum_k=0^[fracn3](-1)^kbinomn+1kbinom2n-3kn=sum_k=[fracn2]^nbinomn+1kbinomkn-k$How find the sum $2sum_k=1^inftysum_i=0^2k-1fracbinom2kicdot B_icdot(m-1)^2k-i2k(2k-1)m^2k-1$How to prove that $sum_k=0^infty binomxx-kcdotbinomxk-x = 1$?prove that $(frac16)^4cdotlim_nrightarrowinftysum_i=4^nbinomi-13(frac56)^i-4=1$Evaluate$sum_n=0^inftybinom3nnx^n$$sum_i=1^ni^k=sum_i=1^kfracc_k+1-ik!i!n^i$. What is known about $c_k_k=0^infty$?$lim_limitssigma_A, sigma_B to infty e^-sigma_A-sigma_B sum_k=0^infty fracsigma_A^kk!cdotfracsigma_B^k+1(k+1)!$Proof for binomial theorem $ (1+x)^a = sum_k=0^infty P_k(a)x^k $Help understanding proof of the following statement $E(Y) = sum_i = 1^infty P(Y geq k)$
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
edited Mar 18 at 12:56
Arthur
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
$endgroup$
add a comment |
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
edited Mar 18 at 12:56
Arthur
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
$endgroup$
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
$begingroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
edited Mar 18 at 12:56
Arthur
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
$endgroup$
$$
sum_i = j^inftyfraci!(i-j)!cdot j!cdot k^i-jcdot (1-k)^j = frac11-k
$$
I take this equal than I do some problem of probabiliti teory. I think that it will be prove by indukcia, but I can't.
probability binomial-coefficients
probability binomial-coefficients
edited Mar 18 at 12:56
Arthur
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
edited Mar 18 at 12:56
Arthur
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
edited Mar 18 at 12:56
Arthur
120k7121206
edited Mar 18 at 12:56
Arthur
120k7121206
edited Mar 18 at 12:56
Arthur
120k7121206
120k7121206
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
asked Mar 18 at 12:53
Andrey KomisarovAndrey Komisarov
726
726
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40
add a comment |
1 Answer
1
active
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votes
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
$endgroup$
add a comment |
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
$endgroup$
add a comment |
$begingroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
$endgroup$
it helps to write $i=j+l$, then divide out $(1-k)^j$ to restate the claim as $sum_lge 0binomj+ljk^l=(1-k)^-j-1$. This is just a special case of the binomial theorem, since the falling Pochhammer symbol $(-j-1)_l=binomj+lj(-1)^l$.
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
answered Mar 18 at 13:00
J.G.J.G.
32.3k23250
32.3k23250
add a comment |
add a comment |
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$begingroup$
I have edited your expression using MathJax. Could you please check it to see if I did any mistakes. And for next time, consider using it yourself; it makes formulas much easier to read.
$endgroup$
– Arthur
Mar 18 at 12:57
$begingroup$
Sorry, I forgot to say that k is probability. So it is 0<k<1.
$endgroup$
– Andrey Komisarov
Mar 18 at 13:00
$begingroup$
@AndreyKomisarov It is not clear what you mean. Please check if it is the following one: $sum_i=1^infty sum_j=i^infty i choose jcdot k^i-jcdot (1-k)^j$.
$endgroup$
– callculus
Mar 18 at 15:56
$begingroup$
@callculus No, j is a constant, the suming is only by i.
$endgroup$
– Andrey Komisarov
Mar 18 at 19:40