The calculated series of $T(n)$ is: $T(n)= {1,3,6,10,15,21,28,36,45,55,66}$ for $n=0$ to $n=10$, how do you...












1














The calculated series of $T(n)$ is: $T(n)= {1,3,6,10,15,21,28,36,45,55,66}$ for $n=0$ to $n=10$, how do you write this using sigma notation?
$T(n)$ contains the functions:





  • $T(n−1) + (n+1)$ for $ngeq1$

  • $T(0) = 1$






sequences-and-series summation







share|cite|improve this question




















  • 1




    You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
    – Batominovski
    Sep 23 '18 at 17:21








  • 1




    These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
    – Matt
    Sep 23 '18 at 17:21










  • for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
    – JScribe
    Sep 23 '18 at 17:33








  • 1




    I hope you realize @Batominovski was being a bit facetious.
    – Don Thousand
    Sep 26 '18 at 23:09
















1














The calculated series of $T(n)$ is: $T(n)= {1,3,6,10,15,21,28,36,45,55,66}$ for $n=0$ to $n=10$, how do you write this using sigma notation?
$T(n)$ contains the functions:





  • $T(n−1) + (n+1)$ for $ngeq1$

  • $T(0) = 1$






sequences-and-series summation







share|cite|improve this question




















  • 1




    You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
    – Batominovski
    Sep 23 '18 at 17:21








  • 1




    These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
    – Matt
    Sep 23 '18 at 17:21










  • for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
    – JScribe
    Sep 23 '18 at 17:33








  • 1




    I hope you realize @Batominovski was being a bit facetious.
    – Don Thousand
    Sep 26 '18 at 23:09














1












1








1







The calculated series of $T(n)$ is: $T(n)= {1,3,6,10,15,21,28,36,45,55,66}$ for $n=0$ to $n=10$, how do you write this using sigma notation?
$T(n)$ contains the functions:





  • $T(n−1) + (n+1)$ for $ngeq1$

  • $T(0) = 1$






sequences-and-series summation







share|cite|improve this question















The calculated series of $T(n)$ is: $T(n)= {1,3,6,10,15,21,28,36,45,55,66}$ for $n=0$ to $n=10$, how do you write this using sigma notation?
$T(n)$ contains the functions:





  • $T(n−1) + (n+1)$ for $ngeq1$

  • $T(0) = 1$






sequences-and-series summation




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 Sep 23 '18 at 18:35









rtybase

10.5k21533




10.5k21533










asked Sep 23 '18 at 17:16









JScribeJScribe

184




184








  • 1




    You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
    – Batominovski
    Sep 23 '18 at 17:21








  • 1




    These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
    – Matt
    Sep 23 '18 at 17:21










  • for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
    – JScribe
    Sep 23 '18 at 17:33








  • 1




    I hope you realize @Batominovski was being a bit facetious.
    – Don Thousand
    Sep 26 '18 at 23:09














  • 1




    You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
    – Batominovski
    Sep 23 '18 at 17:21








  • 1




    These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
    – Matt
    Sep 23 '18 at 17:21










  • for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
    – JScribe
    Sep 23 '18 at 17:33








  • 1




    I hope you realize @Batominovski was being a bit facetious.
    – Don Thousand
    Sep 26 '18 at 23:09








1




1




You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
– Batominovski
Sep 23 '18 at 17:21






You mean something like this: $$T(n)=sum_{k=1}^{frac{(n+1)(n+2)}{2}},1,?$$ I used sigma notation as requested.
– Batominovski
Sep 23 '18 at 17:21






1




1




These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
– Matt
Sep 23 '18 at 17:21




These are triangular numbers. en.wikipedia.org/wiki/Triangular_number answers your question immediately.
– Matt
Sep 23 '18 at 17:21












for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
– JScribe
Sep 23 '18 at 17:33






for Batominovski, how was this arranged? "k=1" is the intial instance, so "((n+1)(n+2))/2" is the nth sequence? What is 1 for? These are indeed triangular numbers but i'm having a hard time understanding what sigma formula would be used for each instance of n from 0 to the nth number since the sequence is not a simple arithmetic or geometric format.
– JScribe
Sep 23 '18 at 17:33






1




1




I hope you realize @Batominovski was being a bit facetious.
– Don Thousand
Sep 26 '18 at 23:09




I hope you realize @Batominovski was being a bit facetious.
– Don Thousand
Sep 26 '18 at 23:09










2 Answers
2






active

oldest

votes


















3














So we have



begin{align}
T(0)&=1\
T(1)&=3\
T(2)&=6\
T(3)&=10\
T(4)&=15\
T(5)&=21\
T(6)&=28\
T(7)&=36\
T(8)&=45\
T(9)&=55\
T(10)&=66\
end{align}
So it appears that
$$ T(n)=sum_{k=0}^n(k+1) $$






share|cite|improve this answer





















  • Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
    – JScribe
    Oct 2 '18 at 3:13



















1














Not to be a smart-arse but the following is another solution:



begin{equation}
S = sum_{t in T(n)} t
end{equation}






share|cite|improve this answer





















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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    So we have



    begin{align}
    T(0)&=1\
    T(1)&=3\
    T(2)&=6\
    T(3)&=10\
    T(4)&=15\
    T(5)&=21\
    T(6)&=28\
    T(7)&=36\
    T(8)&=45\
    T(9)&=55\
    T(10)&=66\
    end{align}
    So it appears that
    $$ T(n)=sum_{k=0}^n(k+1) $$






    share|cite|improve this answer





















    • Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
      – JScribe
      Oct 2 '18 at 3:13
















    3














    So we have



    begin{align}
    T(0)&=1\
    T(1)&=3\
    T(2)&=6\
    T(3)&=10\
    T(4)&=15\
    T(5)&=21\
    T(6)&=28\
    T(7)&=36\
    T(8)&=45\
    T(9)&=55\
    T(10)&=66\
    end{align}
    So it appears that
    $$ T(n)=sum_{k=0}^n(k+1) $$






    share|cite|improve this answer





















    • Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
      – JScribe
      Oct 2 '18 at 3:13














    3












    3








    3






    So we have



    begin{align}
    T(0)&=1\
    T(1)&=3\
    T(2)&=6\
    T(3)&=10\
    T(4)&=15\
    T(5)&=21\
    T(6)&=28\
    T(7)&=36\
    T(8)&=45\
    T(9)&=55\
    T(10)&=66\
    end{align}
    So it appears that
    $$ T(n)=sum_{k=0}^n(k+1) $$






    share|cite|improve this answer












    So we have



    begin{align}
    T(0)&=1\
    T(1)&=3\
    T(2)&=6\
    T(3)&=10\
    T(4)&=15\
    T(5)&=21\
    T(6)&=28\
    T(7)&=36\
    T(8)&=45\
    T(9)&=55\
    T(10)&=66\
    end{align}
    So it appears that
    $$ T(n)=sum_{k=0}^n(k+1) $$







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Sep 23 '18 at 17:49









    John Wayland BalesJohn Wayland Bales

    13.9k21238




    13.9k21238












    • Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
      – JScribe
      Oct 2 '18 at 3:13


















    • Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
      – JScribe
      Oct 2 '18 at 3:13
















    Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
    – JScribe
    Oct 2 '18 at 3:13




    Thank you! that makes a lot of sense! and I found out k is better for sigma than doing multiple n notations.
    – JScribe
    Oct 2 '18 at 3:13











    1














    Not to be a smart-arse but the following is another solution:



    begin{equation}
    S = sum_{t in T(n)} t
    end{equation}






    share|cite|improve this answer


























      1














      Not to be a smart-arse but the following is another solution:



      begin{equation}
      S = sum_{t in T(n)} t
      end{equation}






      share|cite|improve this answer
























        1












        1








        1






        Not to be a smart-arse but the following is another solution:



        begin{equation}
        S = sum_{t in T(n)} t
        end{equation}






        share|cite|improve this answer












        Not to be a smart-arse but the following is another solution:



        begin{equation}
        S = sum_{t in T(n)} t
        end{equation}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 4 at 7:37









        DavidGDavidG

        1,874620




        1,874620






























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