Calculate the sum of the first N terms of the sequence
$begingroup$
$a_n=a_{n-1}displaystyle frac{n+1}{n}$ 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
$endgroup$
add a comment |
$begingroup$
$a_n=a_{n-1}displaystyle frac{n+1}{n}$ 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
$endgroup$
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
add a comment |
$begingroup$
$a_n=a_{n-1}displaystyle frac{n+1}{n}$ 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
$endgroup$
$a_n=a_{n-1}displaystyle frac{n+1}{n}$ 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
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
add a comment |
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
add a comment |
1 Answer
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oldest
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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
$endgroup$
add a comment |
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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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
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
answered Dec 18 '18 at 8:30
HenryHenry
100k481168
100k481168
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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