Summation rules and properties

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I am trying to find the sum of this -



$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$



I know roughly how I am supposed to do. First I distribute the summation across the 3 values.



Then I got stuck $2r$ and $0.3^r$



Both are similar if I understand, so I will only ask for one of them ,



$sum_r=1^100 (2r)$



how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $



Am I right to say -



$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.




summation




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  • Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
    – Shrey Joshi
    Jul 22 at 17:20










  • $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
    – Henry
    Jul 22 at 17:25















up vote
0
down vote

favorite












I am trying to find the sum of this -



$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$



I know roughly how I am supposed to do. First I distribute the summation across the 3 values.



Then I got stuck $2r$ and $0.3^r$



Both are similar if I understand, so I will only ask for one of them ,



$sum_r=1^100 (2r)$



how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $



Am I right to say -



$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.




summation




share|cite|improve this question





















  • Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
    – Shrey Joshi
    Jul 22 at 17:20










  • $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
    – Henry
    Jul 22 at 17:25













up vote
0
down vote

favorite









up vote
0
down vote

favorite











I am trying to find the sum of this -



$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$



I know roughly how I am supposed to do. First I distribute the summation across the 3 values.



Then I got stuck $2r$ and $0.3^r$



Both are similar if I understand, so I will only ask for one of them ,



$sum_r=1^100 (2r)$



how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $



Am I right to say -



$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.




summation




share|cite|improve this question













I am trying to find the sum of this -



$$sum_r=1^100 ( 1 + 2r + 0.3^r ) $$



I know roughly how I am supposed to do. First I distribute the summation across the 3 values.



Then I got stuck $2r$ and $0.3^r$



Both are similar if I understand, so I will only ask for one of them ,



$sum_r=1^100 (2r)$



how am I suppose to use the property to solve this just like $sum_r=1^100 = 100(1) = 100 $



Am I right to say -



$sum_r=1^100 (2r) = 2(1)(100) $ ? But I doubt myself as the r value changes . So I think this is wrong.





summation





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 22 at 17:20
























asked Jul 22 at 17:18









user185692

1045




1045











  • Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
    – Shrey Joshi
    Jul 22 at 17:20










  • $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
    – Henry
    Jul 22 at 17:25

















  • Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
    – Shrey Joshi
    Jul 22 at 17:20










  • $sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
    – Henry
    Jul 22 at 17:25
















Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
– Shrey Joshi
Jul 22 at 17:20




Try separating the sum into sums of $1$, $2r$, and $0.3^r$ then solving each independently.
– Shrey Joshi
Jul 22 at 17:20












$sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
– Henry
Jul 22 at 17:25





$sum_r=1^100 (2r) = (1+2+cdots+99+100) + (100+99+cdots+2+1) $ $= (1+100)+(2+99)+cdots+(99+2)+(100+1) = 100 times 101$
– Henry
Jul 22 at 17:25











2 Answers
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up vote
3
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accepted










You should try breaking up the summation into parts so that you can evaluate them easily



$displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$



$=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$



$displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$



$displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$



$=displaystyle 100+10100+1.428 $



$=10201.4285$



EDIT:



You should look up geometric series to understand the summation of $(0,3)^r $






share|cite|improve this answer




























    up vote
    2
    down vote













    Hint:
    $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$



    Solution:




    $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
    $$left(sum_r=1^1001right)=100$$
    $$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
    Since $1+2+3+...+r=r(r+1)/2$
    $$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
    The sum is about $10201.428$







    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

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      active

      oldest

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      up vote
      3
      down vote



      accepted










      You should try breaking up the summation into parts so that you can evaluate them easily



      $displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$



      $=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$



      $displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$



      $displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$



      $=displaystyle 100+10100+1.428 $



      $=10201.4285$



      EDIT:



      You should look up geometric series to understand the summation of $(0,3)^r $






      share|cite|improve this answer

























        up vote
        3
        down vote



        accepted










        You should try breaking up the summation into parts so that you can evaluate them easily



        $displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$



        $=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$



        $displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$



        $displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$



        $=displaystyle 100+10100+1.428 $



        $=10201.4285$



        EDIT:



        You should look up geometric series to understand the summation of $(0,3)^r $






        share|cite|improve this answer























          up vote
          3
          down vote



          accepted







          up vote
          3
          down vote



          accepted






          You should try breaking up the summation into parts so that you can evaluate them easily



          $displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$



          $=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$



          $displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$



          $displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$



          $=displaystyle 100+10100+1.428 $



          $=10201.4285$



          EDIT:



          You should look up geometric series to understand the summation of $(0,3)^r $






          share|cite|improve this answer













          You should try breaking up the summation into parts so that you can evaluate them easily



          $displaystyle sum_r=1^100 ( 1 + 2r + 0.3^r )$



          $=displaystyle sum_r=1^1001+sum_r=1^1002r+sum_r=1^100(0.3)^r$



          $displaystyle = 100+2sum_r=1^100r+sum_r=1^100(0.3)^r$



          $displaystyle= 100+2cdotfrac100(101)2+ frac1-(0.3)^1001-0.3$



          $=displaystyle 100+10100+1.428 $



          $=10201.4285$



          EDIT:



          You should look up geometric series to understand the summation of $(0,3)^r $







          share|cite|improve this answer













          share|cite|improve this answer



          share|cite|improve this answer











          answered Jul 22 at 17:24









          The Integrator

          3,0262725




          3,0262725




















              up vote
              2
              down vote













              Hint:
              $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$



              Solution:




              $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
              $$left(sum_r=1^1001right)=100$$
              $$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
              Since $1+2+3+...+r=r(r+1)/2$
              $$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
              The sum is about $10201.428$







              share|cite|improve this answer

























                up vote
                2
                down vote













                Hint:
                $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$



                Solution:




                $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
                $$left(sum_r=1^1001right)=100$$
                $$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
                Since $1+2+3+...+r=r(r+1)/2$
                $$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
                The sum is about $10201.428$







                share|cite|improve this answer























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Hint:
                  $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$



                  Solution:




                  $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
                  $$left(sum_r=1^1001right)=100$$
                  $$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
                  Since $1+2+3+...+r=r(r+1)/2$
                  $$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
                  The sum is about $10201.428$







                  share|cite|improve this answer













                  Hint:
                  $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$



                  Solution:




                  $$sum_r=1^1001+2r+0.3^r=left(sum_r=1^1001right)+left(sum_r=1^1002rright)+left(sum_r=1^1000.3^rright)$$
                  $$left(sum_r=1^1001right)=100$$
                  $$left(sum_r=1^1002rright)=2sum_r=1^100r=100(101)=10100$$
                  Since $1+2+3+...+r=r(r+1)/2$
                  $$left(sum_r=1^1000.3^rright)=frac1-0.3^1001-0.3$$
                  The sum is about $10201.428$








                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 22 at 17:29









                  Shrey Joshi

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