what geomatric series formula is used?
$begingroup$
trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work
sequences-and-series geometry analytic-geometry geometric-construction
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
$endgroup$
add a comment |
$begingroup$
trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work
sequences-and-series geometry analytic-geometry geometric-construction
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
$endgroup$
add a comment |
$begingroup$
trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work
sequences-and-series geometry analytic-geometry geometric-construction
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
$endgroup$
trying to follow a solution related to geometric series, but not sure what formula is used here. Any pointer is appreciated.Image here, can't embed image yet. I do try to plug in the Sn formula, but didn't get it to work
sequences-and-series geometry analytic-geometry geometric-construction
sequences-and-series geometry analytic-geometry geometric-construction
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
edited Dec 13 '18 at 3:11
André 3000
12.6k22243
12.6k22243
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
asked Dec 13 '18 at 2:30
MaxfieldMaxfield
424
424
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$sum_{k=0}^n 3^{n-k} cdot 2^k = sum_{k=0}^n 3^{n} cdot 3^{-k} cdot 2^k =sum_{k=0}^n 3^{n} cdot frac{2^k}{3^k} = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k$$
This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by
$$sum_{k=0}^n ar^k = frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $3$ on the top and bottom:
$$frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.
$$frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3} = frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = frac{3^{n+1}}{3^{n+1}} - frac{2^{n+1}}{3^{n+1}} = frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with
$$frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2} = frac{3^{n+1} cdot frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = 3^{n+1} - 2^{n+1}$$
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
add a comment |
$begingroup$
Sum of first n terms of a geometric series is given by:
$$S_n = frac{a(r^n-1)}{r-1}.$$
Here $a$ is the first term of the series and $r$ is the common ratio.
In your question, $r = frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
$endgroup$
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
add a comment |
$begingroup$
One gets $T(2n)=3^nsum_{k=0}^n(frac23)^k=3^n(frac{1-(frac23)^{n+1}}{1-frac23})=3^{n+1}-2^{n+1}$.
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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$begingroup$
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$sum_{k=0}^n 3^{n-k} cdot 2^k = sum_{k=0}^n 3^{n} cdot 3^{-k} cdot 2^k =sum_{k=0}^n 3^{n} cdot frac{2^k}{3^k} = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k$$
This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by
$$sum_{k=0}^n ar^k = frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $3$ on the top and bottom:
$$frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.
$$frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3} = frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = frac{3^{n+1}}{3^{n+1}} - frac{2^{n+1}}{3^{n+1}} = frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with
$$frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2} = frac{3^{n+1} cdot frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = 3^{n+1} - 2^{n+1}$$
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
add a comment |
$begingroup$
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$sum_{k=0}^n 3^{n-k} cdot 2^k = sum_{k=0}^n 3^{n} cdot 3^{-k} cdot 2^k =sum_{k=0}^n 3^{n} cdot frac{2^k}{3^k} = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k$$
This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by
$$sum_{k=0}^n ar^k = frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $3$ on the top and bottom:
$$frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.
$$frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3} = frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = frac{3^{n+1}}{3^{n+1}} - frac{2^{n+1}}{3^{n+1}} = frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with
$$frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2} = frac{3^{n+1} cdot frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = 3^{n+1} - 2^{n+1}$$
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
add a comment |
$begingroup$
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$sum_{k=0}^n 3^{n-k} cdot 2^k = sum_{k=0}^n 3^{n} cdot 3^{-k} cdot 2^k =sum_{k=0}^n 3^{n} cdot frac{2^k}{3^k} = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k$$
This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by
$$sum_{k=0}^n ar^k = frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $3$ on the top and bottom:
$$frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.
$$frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3} = frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = frac{3^{n+1}}{3^{n+1}} - frac{2^{n+1}}{3^{n+1}} = frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with
$$frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2} = frac{3^{n+1} cdot frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = 3^{n+1} - 2^{n+1}$$
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
$endgroup$
Notice:
$$3^n + 2(3^{n-1}) + 2^2 (3^{n-2}) + ... 2^{n-1}(3) + 2^n$$
can be effectively rewritten as
$$sum_{k=0}^n 3^{n-k} cdot 2^k = sum_{k=0}^n 3^{n} cdot 3^{-k} cdot 2^k =sum_{k=0}^n 3^{n} cdot frac{2^k}{3^k} = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k$$
This is a finite geometric series, then, with ratio $r=2/3$ and first term $2^03^n = 3^n$. The sum of a finite geometric series $a + ar + ar^2 + ... +ar^n$ is given by
$$sum_{k=0}^n ar^k = frac{a(r^{n+1} - 1)}{r-1}$$
Thus,
$$sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = frac{3^n((2/3)^{n+1} - 1)}{2/3-1}$$
This is essentially the same expression as presented to you, just needing some manipulations to make the equivalence clear.
Multiply by $3$ on the top and bottom:
$$frac{3^n((2/3)^{n+1} - 1)}{2/3-1} = frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3}$$
Multiply by $-1$ on the top and bottom. We distribute this into the $()$ on top to reverse the order.
$$frac{3^{n+1} cdot ((2/3)^{n+1} - 1)}{2 - 3} = frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2}$$
Next:
$$1 - (2/3)^{n+1} = frac{3^{n+1}}{3^{n+1}} - frac{2^{n+1}}{3^{n+1}} = frac{3^{n+1} - 2^{n+1}}{3^{n+1}}$$
The denominator cancels owing to the outside $3^{n+1}$ term, and of course $3-2=1$, leaving us with
$$frac{3^{n+1} cdot (1 - (2/3)^{n+1})}{3 - 2} = frac{3^{n+1} cdot frac{3^{n+1} - 2^{n+1}}{3^{n+1}}}{1} = 3^{n+1} - 2^{n+1}$$
Thus,
$$3^n + 2(3^{n-1}) + ... +2^{n-1}(3) + 2^n = sum_{k=0}^n 3^n cdot left( frac{2}{3} right)^k = 3^{n+1} - 2^{n+1}$$
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
edited Dec 13 '18 at 3:41
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
answered Dec 13 '18 at 2:59
Eevee TrainerEevee Trainer
5,9781936
5,9781936
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
add a comment |
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
It looks like your sum has an extra $(2/3)^k$ in it.
$endgroup$
– Michael Burr
Dec 13 '18 at 3:32
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Where, exactly?
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:36
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Oh, I see. I'll try to figure out how to resolve it.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:37
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
$begingroup$
Okay, I believe I fixed it. It didn't really affect my approach, just more how it was being written in the summation. I suck at writing summations when they're a bit awkward like the one givenn.
$endgroup$
– Eevee Trainer
Dec 13 '18 at 3:42
add a comment |
$begingroup$
Sum of first n terms of a geometric series is given by:
$$S_n = frac{a(r^n-1)}{r-1}.$$
Here $a$ is the first term of the series and $r$ is the common ratio.
In your question, $r = frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
$endgroup$
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
add a comment |
$begingroup$
Sum of first n terms of a geometric series is given by:
$$S_n = frac{a(r^n-1)}{r-1}.$$
Here $a$ is the first term of the series and $r$ is the common ratio.
In your question, $r = frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
$endgroup$
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
add a comment |
$begingroup$
Sum of first n terms of a geometric series is given by:
$$S_n = frac{a(r^n-1)}{r-1}.$$
Here $a$ is the first term of the series and $r$ is the common ratio.
In your question, $r = frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
$endgroup$
Sum of first n terms of a geometric series is given by:
$$S_n = frac{a(r^n-1)}{r-1}.$$
Here $a$ is the first term of the series and $r$ is the common ratio.
In your question, $r = frac{2}{3}$, $a=3^n$ and the number of terms is $n+1$ (you probably messed up in this part).
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
edited Dec 13 '18 at 2:52
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
answered Dec 13 '18 at 2:44
Nutan NepalNutan Nepal
505
505
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
add a comment |
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
yes, did you notice that it was stated as $T(2^n)$ = ..., I do try to plug in the $S_n$ formula, but didn't get it to work
$endgroup$
– Maxfield
Dec 13 '18 at 2:45
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
Perhaps it would be helpful to identify $a$ and $r$ in the given problem.
$endgroup$
– Michael Burr
Dec 13 '18 at 2:46
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
as stated in the solution, r is 2/3 and, as a, which I can't find. but it states as T($2^n$) instead of $S_n$ though.
$endgroup$
– Maxfield
Dec 13 '18 at 2:48
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
$begingroup$
It doesn't matter what it is stated as. You can just calculate the sum of the RHS. (Also, see edit)
$endgroup$
– Nutan Nepal
Dec 13 '18 at 2:54
add a comment |
$begingroup$
One gets $T(2n)=3^nsum_{k=0}^n(frac23)^k=3^n(frac{1-(frac23)^{n+1}}{1-frac23})=3^{n+1}-2^{n+1}$.
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
One gets $T(2n)=3^nsum_{k=0}^n(frac23)^k=3^n(frac{1-(frac23)^{n+1}}{1-frac23})=3^{n+1}-2^{n+1}$.
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
$endgroup$
add a comment |
$begingroup$
One gets $T(2n)=3^nsum_{k=0}^n(frac23)^k=3^n(frac{1-(frac23)^{n+1}}{1-frac23})=3^{n+1}-2^{n+1}$.
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
$endgroup$
One gets $T(2n)=3^nsum_{k=0}^n(frac23)^k=3^n(frac{1-(frac23)^{n+1}}{1-frac23})=3^{n+1}-2^{n+1}$.
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
answered Dec 13 '18 at 3:13
Chris CusterChris Custer
13.2k3827
13.2k3827
add a comment |
add a comment |
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