Calculate the sum of the first N terms of the sequence












5












$begingroup$


$a_n=a_{n-1}displaystyle frac{n+1}{n}$ if $n > 1$



$a_n=1$ if $n=1$



I'm not too sure where to start here. This is part of a review for a class and I can't really seem to remember what we're reviewing. The first 5 values are...



$a_1 = 1,a_2=1.5,a_3=2,a_4=2.5,a_5=3$



Sums of these to each point....



$a_1=1, a_2=2.5, a_3=4.5, a_4=7, a_5=10$



It doesn't seem like it should be too tricky to figure out how to get a formula for a sum of the first N terms, since each term seems to just increase by 0.5 every team, I just haven't done this for a while and am a little rusty. Any pointers would be greatly appreciated!






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know how to sum an Arithmetic Progression?
    $endgroup$
    – Rijul Saini
    Sep 5 '12 at 15:58












  • $begingroup$
    Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:02








  • 1




    $begingroup$
    @Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
    $endgroup$
    – Ross Millikan
    Sep 5 '12 at 16:08










  • $begingroup$
    Yeah I see what you mean. Thanks!
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:19










  • $begingroup$
    The trick is to show that $a_i$ is an arithmetic progression.
    $endgroup$
    – Thomas Andrews
    Sep 5 '12 at 16:53
















5












$begingroup$


$a_n=a_{n-1}displaystyle frac{n+1}{n}$ if $n > 1$



$a_n=1$ if $n=1$



I'm not too sure where to start here. This is part of a review for a class and I can't really seem to remember what we're reviewing. The first 5 values are...



$a_1 = 1,a_2=1.5,a_3=2,a_4=2.5,a_5=3$



Sums of these to each point....



$a_1=1, a_2=2.5, a_3=4.5, a_4=7, a_5=10$



It doesn't seem like it should be too tricky to figure out how to get a formula for a sum of the first N terms, since each term seems to just increase by 0.5 every team, I just haven't done this for a while and am a little rusty. Any pointers would be greatly appreciated!






sequences-and-series







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Do you know how to sum an Arithmetic Progression?
    $endgroup$
    – Rijul Saini
    Sep 5 '12 at 15:58












  • $begingroup$
    Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:02








  • 1




    $begingroup$
    @Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
    $endgroup$
    – Ross Millikan
    Sep 5 '12 at 16:08










  • $begingroup$
    Yeah I see what you mean. Thanks!
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:19










  • $begingroup$
    The trick is to show that $a_i$ is an arithmetic progression.
    $endgroup$
    – Thomas Andrews
    Sep 5 '12 at 16:53














5












5








5





$begingroup$


$a_n=a_{n-1}displaystyle frac{n+1}{n}$ if $n > 1$



$a_n=1$ if $n=1$



I'm not too sure where to start here. This is part of a review for a class and I can't really seem to remember what we're reviewing. The first 5 values are...



$a_1 = 1,a_2=1.5,a_3=2,a_4=2.5,a_5=3$



Sums of these to each point....



$a_1=1, a_2=2.5, a_3=4.5, a_4=7, a_5=10$



It doesn't seem like it should be too tricky to figure out how to get a formula for a sum of the first N terms, since each term seems to just increase by 0.5 every team, I just haven't done this for a while and am a little rusty. Any pointers would be greatly appreciated!






sequences-and-series







share|cite|improve this question











$endgroup$




$a_n=a_{n-1}displaystyle frac{n+1}{n}$ if $n > 1$



$a_n=1$ if $n=1$



I'm not too sure where to start here. This is part of a review for a class and I can't really seem to remember what we're reviewing. The first 5 values are...



$a_1 = 1,a_2=1.5,a_3=2,a_4=2.5,a_5=3$



Sums of these to each point....



$a_1=1, a_2=2.5, a_3=4.5, a_4=7, a_5=10$



It doesn't seem like it should be too tricky to figure out how to get a formula for a sum of the first N terms, since each term seems to just increase by 0.5 every team, I just haven't done this for a while and am a little rusty. Any pointers would be greatly appreciated!






sequences-and-series




sequences-and-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Sep 5 '12 at 16:04









Cameron Buie

85.5k772158




85.5k772158










asked Sep 5 '12 at 15:55









HoserHoser

215312




215312








  • 1




    $begingroup$
    Do you know how to sum an Arithmetic Progression?
    $endgroup$
    – Rijul Saini
    Sep 5 '12 at 15:58












  • $begingroup$
    Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:02








  • 1




    $begingroup$
    @Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
    $endgroup$
    – Ross Millikan
    Sep 5 '12 at 16:08










  • $begingroup$
    Yeah I see what you mean. Thanks!
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:19










  • $begingroup$
    The trick is to show that $a_i$ is an arithmetic progression.
    $endgroup$
    – Thomas Andrews
    Sep 5 '12 at 16:53














  • 1




    $begingroup$
    Do you know how to sum an Arithmetic Progression?
    $endgroup$
    – Rijul Saini
    Sep 5 '12 at 15:58












  • $begingroup$
    Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:02








  • 1




    $begingroup$
    @Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
    $endgroup$
    – Ross Millikan
    Sep 5 '12 at 16:08










  • $begingroup$
    Yeah I see what you mean. Thanks!
    $endgroup$
    – Hoser
    Sep 5 '12 at 16:19










  • $begingroup$
    The trick is to show that $a_i$ is an arithmetic progression.
    $endgroup$
    – Thomas Andrews
    Sep 5 '12 at 16:53








1




1




$begingroup$
Do you know how to sum an Arithmetic Progression?
$endgroup$
– Rijul Saini
Sep 5 '12 at 15:58






$begingroup$
Do you know how to sum an Arithmetic Progression?
$endgroup$
– Rijul Saini
Sep 5 '12 at 15:58














$begingroup$
Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
$endgroup$
– Hoser
Sep 5 '12 at 16:02






$begingroup$
Got it! Thank you. Answer seems to be $n/2(a_1 + a_{n})$
$endgroup$
– Hoser
Sep 5 '12 at 16:02






1




1




$begingroup$
@Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
$endgroup$
– Ross Millikan
Sep 5 '12 at 16:08




$begingroup$
@Hoser: correct, but it is better to write $n(a_1+a_n)/2$ so it is obvious that the $(a_1+a_n)$ is in the numerator.
$endgroup$
– Ross Millikan
Sep 5 '12 at 16:08












$begingroup$
Yeah I see what you mean. Thanks!
$endgroup$
– Hoser
Sep 5 '12 at 16:19




$begingroup$
Yeah I see what you mean. Thanks!
$endgroup$
– Hoser
Sep 5 '12 at 16:19












$begingroup$
The trick is to show that $a_i$ is an arithmetic progression.
$endgroup$
– Thomas Andrews
Sep 5 '12 at 16:53




$begingroup$
The trick is to show that $a_i$ is an arithmetic progression.
$endgroup$
– Thomas Andrews
Sep 5 '12 at 16:53










1 Answer
1






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oldest

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1












$begingroup$

Just to give a complete answer, the pattern for each term appears to be $a_n=frac{n}{2}+frac12$



To prove it, this correctly gives $a_1=1$ and $a_{n-1}frac{n+1}{n}= left(frac{n-1}{2}+frac{1}{2}right)frac{n+1}{n}=frac{n+1}{2}=frac{n}{2}+frac12$ so the hypothesis is true by induction



As you say, the sum of the first $n$ terms is $frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $frac{nleft(1+frac{n}{2}+frac12right)}{2} = frac{nleft(n+3right)}{4}$ and for example gives the sum of the first five terms as $frac{5 times 8}{4} = 10$ as you found






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    1 Answer
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    $begingroup$

    Just to give a complete answer, the pattern for each term appears to be $a_n=frac{n}{2}+frac12$



    To prove it, this correctly gives $a_1=1$ and $a_{n-1}frac{n+1}{n}= left(frac{n-1}{2}+frac{1}{2}right)frac{n+1}{n}=frac{n+1}{2}=frac{n}{2}+frac12$ so the hypothesis is true by induction



    As you say, the sum of the first $n$ terms is $frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $frac{nleft(1+frac{n}{2}+frac12right)}{2} = frac{nleft(n+3right)}{4}$ and for example gives the sum of the first five terms as $frac{5 times 8}{4} = 10$ as you found






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Just to give a complete answer, the pattern for each term appears to be $a_n=frac{n}{2}+frac12$



      To prove it, this correctly gives $a_1=1$ and $a_{n-1}frac{n+1}{n}= left(frac{n-1}{2}+frac{1}{2}right)frac{n+1}{n}=frac{n+1}{2}=frac{n}{2}+frac12$ so the hypothesis is true by induction



      As you say, the sum of the first $n$ terms is $frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $frac{nleft(1+frac{n}{2}+frac12right)}{2} = frac{nleft(n+3right)}{4}$ and for example gives the sum of the first five terms as $frac{5 times 8}{4} = 10$ as you found






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Just to give a complete answer, the pattern for each term appears to be $a_n=frac{n}{2}+frac12$



        To prove it, this correctly gives $a_1=1$ and $a_{n-1}frac{n+1}{n}= left(frac{n-1}{2}+frac{1}{2}right)frac{n+1}{n}=frac{n+1}{2}=frac{n}{2}+frac12$ so the hypothesis is true by induction



        As you say, the sum of the first $n$ terms is $frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $frac{nleft(1+frac{n}{2}+frac12right)}{2} = frac{nleft(n+3right)}{4}$ and for example gives the sum of the first five terms as $frac{5 times 8}{4} = 10$ as you found






        share|cite|improve this answer









        $endgroup$



        Just to give a complete answer, the pattern for each term appears to be $a_n=frac{n}{2}+frac12$



        To prove it, this correctly gives $a_1=1$ and $a_{n-1}frac{n+1}{n}= left(frac{n-1}{2}+frac{1}{2}right)frac{n+1}{n}=frac{n+1}{2}=frac{n}{2}+frac12$ so the hypothesis is true by induction



        As you say, the sum of the first $n$ terms is $frac{n(a_1+a_n)}{2}$, since this is an arithmetic progression. This is $frac{nleft(1+frac{n}{2}+frac12right)}{2} = frac{nleft(n+3right)}{4}$ and for example gives the sum of the first five terms as $frac{5 times 8}{4} = 10$ as you found







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 18 '18 at 8:30









        HenryHenry

        100k481168




        100k481168






























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