Geometric series starting from -n
How do I find the following series:
$$sum^n_{k = -n}ar^k$$
I'm not sure how to manipulate the standard geometric series to get the desired result.
Any direction would be appreciated.
sequences-and-series
edited Nov 20 at 1:59
amWhy
191k28224439
asked Nov 20 at 1:57
51n84d
383
add a comment |
How do I find the following series:
$$sum^n_{k = -n}ar^k$$
I'm not sure how to manipulate the standard geometric series to get the desired result.
Any direction would be appreciated.
sequences-and-series
edited Nov 20 at 1:59
amWhy
191k28224439
asked Nov 20 at 1:57
51n84d
383
2
Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
2
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00
add a comment |
How do I find the following series:
$$sum^n_{k = -n}ar^k$$
I'm not sure how to manipulate the standard geometric series to get the desired result.
Any direction would be appreciated.
sequences-and-series
edited Nov 20 at 1:59
amWhy
191k28224439
asked Nov 20 at 1:57
51n84d
383
How do I find the following series:
$$sum^n_{k = -n}ar^k$$
I'm not sure how to manipulate the standard geometric series to get the desired result.
Any direction would be appreciated.
sequences-and-series
sequences-and-series
edited Nov 20 at 1:59
amWhy
191k28224439
asked Nov 20 at 1:57
51n84d
383
edited Nov 20 at 1:59
amWhy
191k28224439
asked Nov 20 at 1:57
51n84d
383
edited Nov 20 at 1:59
amWhy
191k28224439
edited Nov 20 at 1:59
amWhy
191k28224439
edited Nov 20 at 1:59
amWhy
191k28224439
191k28224439
asked Nov 20 at 1:57
51n84d
383
asked Nov 20 at 1:57
51n84d
383
asked Nov 20 at 1:57
51n84d
383
383
2
Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
2
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00
add a comment |
2
Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
2
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00
2
2
Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
2
2
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00
add a comment |
1 Answer
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You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$
answered Nov 20 at 2:06
Ross Millikan
291k23196370
add a comment |
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1 Answer
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You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$
answered Nov 20 at 2:06
Ross Millikan
291k23196370
add a comment |
You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$
answered Nov 20 at 2:06
Ross Millikan
291k23196370
add a comment |
You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$
answered Nov 20 at 2:06
Ross Millikan
291k23196370
You can write $$sum^n_{k = -n}ar^k=sum^n_{k = -n}ar^{k+n}r^{-n}=r^{-n}sum^{2n}_{m = 0}ar^m$$
by defining $m=k+n$
answered Nov 20 at 2:06
Ross Millikan
291k23196370
answered Nov 20 at 2:06
Ross Millikan
291k23196370
answered Nov 20 at 2:06
Ross Millikan
291k23196370
answered Nov 20 at 2:06
Ross Millikan
291k23196370
291k23196370
add a comment |
add a comment |
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Hint: $sumlimits_{k=-n}^n f(k) = left(sumlimits_{k=-n}^{-1}f(k)right)+left(sumlimits_{k=0}^nf(k)right)$ and $sumlimits_{k=-n}^{-1}f(k)=sumlimits_{k=1}^n f(-k)$
– JMoravitz
Nov 20 at 1:58
2
Alternatively $sumlimits_{k=-n}^n f(k) = sumlimits_{k=0}^{2n} f(k-n)$
– JMoravitz
Nov 20 at 2:00