Can $sum_{i=0}^{z-n}frac{p_1^i}{i!}frac{p_{2}^{z-n-i}}{(z-n-i)!}$ be simplified? [on hold]












-1














Can the following expression



$$sum_{i=0}^{z-n}frac{p_1^i}{i!}frac{p_{2}^{z-n-i}}{(z-n-i)!}$$



where $z$ is integer, be simplified (i.e., to loss the sum)?






real-analysis calculus sequences-and-series summation







share|cite|improve this question















put on hold as off-topic by amWhy, Adrian Keister, metamorphy, Leucippus, stressed out 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Adrian Keister, metamorphy, Leucippus, stressed out

If this question can be reworded to fit the rules in the help center, please edit the question.













  • what is $z$ ? integer, real, ..
    – G Cab
    Jan 3 at 21:45






  • 2




    Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
    – robjohn♦
    Jan 3 at 21:46






  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jan 3 at 21:49










  • Thanks @GCab, yes z is an integer.
    – Y.L
    Jan 3 at 21:50






  • 1




    This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
    – Bernard
    Jan 3 at 21:54
















-1














Can the following expression



$$sum_{i=0}^{z-n}frac{p_1^i}{i!}frac{p_{2}^{z-n-i}}{(z-n-i)!}$$



where $z$ is integer, be simplified (i.e., to loss the sum)?






real-analysis calculus sequences-and-series summation







share|cite|improve this question















put on hold as off-topic by amWhy, Adrian Keister, metamorphy, Leucippus, stressed out 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Adrian Keister, metamorphy, Leucippus, stressed out

If this question can be reworded to fit the rules in the help center, please edit the question.













  • what is $z$ ? integer, real, ..
    – G Cab
    Jan 3 at 21:45






  • 2




    Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
    – robjohn♦
    Jan 3 at 21:46






  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jan 3 at 21:49










  • Thanks @GCab, yes z is an integer.
    – Y.L
    Jan 3 at 21:50






  • 1




    This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
    – Bernard
    Jan 3 at 21:54














-1












-1








-1







Can the following expression



$$sum_{i=0}^{z-n}frac{p_1^i}{i!}frac{p_{2}^{z-n-i}}{(z-n-i)!}$$



where $z$ is integer, be simplified (i.e., to loss the sum)?






real-analysis calculus sequences-and-series summation







share|cite|improve this question















Can the following expression



$$sum_{i=0}^{z-n}frac{p_1^i}{i!}frac{p_{2}^{z-n-i}}{(z-n-i)!}$$



where $z$ is integer, be simplified (i.e., to loss the sum)?






real-analysis calculus sequences-and-series summation




real-analysis calculus sequences-and-series summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 3 at 21:53









amWhy

192k28224439




192k28224439










asked Jan 3 at 21:40









Y.L

577




577




put on hold as off-topic by amWhy, Adrian Keister, metamorphy, Leucippus, stressed out 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Adrian Keister, metamorphy, Leucippus, stressed out

If this question can be reworded to fit the rules in the help center, please edit the question.




put on hold as off-topic by amWhy, Adrian Keister, metamorphy, Leucippus, stressed out 2 days ago


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – amWhy, Adrian Keister, metamorphy, Leucippus, stressed out

If this question can be reworded to fit the rules in the help center, please edit the question.












  • what is $z$ ? integer, real, ..
    – G Cab
    Jan 3 at 21:45






  • 2




    Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
    – robjohn♦
    Jan 3 at 21:46






  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jan 3 at 21:49










  • Thanks @GCab, yes z is an integer.
    – Y.L
    Jan 3 at 21:50






  • 1




    This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
    – Bernard
    Jan 3 at 21:54


















  • what is $z$ ? integer, real, ..
    – G Cab
    Jan 3 at 21:45






  • 2




    Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
    – robjohn♦
    Jan 3 at 21:46






  • 1




    Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
    – robjohn♦
    Jan 3 at 21:49










  • Thanks @GCab, yes z is an integer.
    – Y.L
    Jan 3 at 21:50






  • 1




    This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
    – Bernard
    Jan 3 at 21:54
















what is $z$ ? integer, real, ..
– G Cab
Jan 3 at 21:45




what is $z$ ? integer, real, ..
– G Cab
Jan 3 at 21:45




2




2




Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
– robjohn♦
Jan 3 at 21:46




Try multiplying by $(z-n)!$ and consider the Binomial Theorem.
– robjohn♦
Jan 3 at 21:46




1




1




Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn♦
Jan 3 at 21:49




Please give more context. Providing context not only assures that this is not simply copied from a homework assignment, but also allows answers to be better directed at where the problem lies and to be within the proper scope. Please avoid "I have no clue" questions. Defining keywords and trying a simpler, similar problem often helps.
– robjohn♦
Jan 3 at 21:49












Thanks @GCab, yes z is an integer.
– Y.L
Jan 3 at 21:50




Thanks @GCab, yes z is an integer.
– Y.L
Jan 3 at 21:50




1




1




This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
– Bernard
Jan 3 at 21:54




This is just the binomial expansion of $(p_1+p_2)^{z-n}$.
– Bernard
Jan 3 at 21:54










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