How do I get $sqrt{r}(r-1) frac{r^{3n/2}-1}{r^{3/2}-1}$?
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I am asked to compute $int_1^2 sqrt x$ from the definition of integral. From my textbook:
Choose $r= 2^{1/n} > 1$ and let $N_n$ be the grid with division points
$1 = r^0 < r <r^2< cdots <r^{n-1} <r^{n} =2$.
The longest of the intervals determined by this grid is the last, so that $d(N_n) = 2-r^{n-1} = 2(1 - frac{1}{r})$. Since $lim_{ n to infty} r = lim_{n to infty} 2^{1/n} = 1, lim_{n to infty} d(N_n) = 0$, and
$S(N_n)= sqrt{r}(r-1) + sqrt{r^2}(r^2 - r) + cdots + sqrt{r^n}(r^{n}-r^{n-1})$ (1)
= $ sqrt{r}(r-1)(1 + rsqrt{r} + [rsqrt{r}]^2 + cdots + [rsqrt{r}]^{n-1})$ (2)
= $sqrt{r}(r-1) frac{r^{frac{3n}{2}}-1}{r^{frac{3}{2}}-1}$ (3)
I wanted to know how I get from (2) to (3)?
calculus real-analysis multivariable-calculus
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
$endgroup$
add a comment |
$begingroup$
I am asked to compute $int_1^2 sqrt x$ from the definition of integral. From my textbook:
Choose $r= 2^{1/n} > 1$ and let $N_n$ be the grid with division points
$1 = r^0 < r <r^2< cdots <r^{n-1} <r^{n} =2$.
The longest of the intervals determined by this grid is the last, so that $d(N_n) = 2-r^{n-1} = 2(1 - frac{1}{r})$. Since $lim_{ n to infty} r = lim_{n to infty} 2^{1/n} = 1, lim_{n to infty} d(N_n) = 0$, and
$S(N_n)= sqrt{r}(r-1) + sqrt{r^2}(r^2 - r) + cdots + sqrt{r^n}(r^{n}-r^{n-1})$ (1)
= $ sqrt{r}(r-1)(1 + rsqrt{r} + [rsqrt{r}]^2 + cdots + [rsqrt{r}]^{n-1})$ (2)
= $sqrt{r}(r-1) frac{r^{frac{3n}{2}}-1}{r^{frac{3}{2}}-1}$ (3)
I wanted to know how I get from (2) to (3)?
calculus real-analysis multivariable-calculus
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
$endgroup$
2
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
1
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09
add a comment |
$begingroup$
I am asked to compute $int_1^2 sqrt x$ from the definition of integral. From my textbook:
Choose $r= 2^{1/n} > 1$ and let $N_n$ be the grid with division points
$1 = r^0 < r <r^2< cdots <r^{n-1} <r^{n} =2$.
The longest of the intervals determined by this grid is the last, so that $d(N_n) = 2-r^{n-1} = 2(1 - frac{1}{r})$. Since $lim_{ n to infty} r = lim_{n to infty} 2^{1/n} = 1, lim_{n to infty} d(N_n) = 0$, and
$S(N_n)= sqrt{r}(r-1) + sqrt{r^2}(r^2 - r) + cdots + sqrt{r^n}(r^{n}-r^{n-1})$ (1)
= $ sqrt{r}(r-1)(1 + rsqrt{r} + [rsqrt{r}]^2 + cdots + [rsqrt{r}]^{n-1})$ (2)
= $sqrt{r}(r-1) frac{r^{frac{3n}{2}}-1}{r^{frac{3}{2}}-1}$ (3)
I wanted to know how I get from (2) to (3)?
calculus real-analysis multivariable-calculus
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
$endgroup$
I am asked to compute $int_1^2 sqrt x$ from the definition of integral. From my textbook:
Choose $r= 2^{1/n} > 1$ and let $N_n$ be the grid with division points
$1 = r^0 < r <r^2< cdots <r^{n-1} <r^{n} =2$.
The longest of the intervals determined by this grid is the last, so that $d(N_n) = 2-r^{n-1} = 2(1 - frac{1}{r})$. Since $lim_{ n to infty} r = lim_{n to infty} 2^{1/n} = 1, lim_{n to infty} d(N_n) = 0$, and
$S(N_n)= sqrt{r}(r-1) + sqrt{r^2}(r^2 - r) + cdots + sqrt{r^n}(r^{n}-r^{n-1})$ (1)
= $ sqrt{r}(r-1)(1 + rsqrt{r} + [rsqrt{r}]^2 + cdots + [rsqrt{r}]^{n-1})$ (2)
= $sqrt{r}(r-1) frac{r^{frac{3n}{2}}-1}{r^{frac{3}{2}}-1}$ (3)
I wanted to know how I get from (2) to (3)?
calculus real-analysis multivariable-calculus
calculus real-analysis multivariable-calculus
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
edited Nov 28 '18 at 5:58
Henning Makholm
239k17304541
239k17304541
asked Nov 28 '18 at 5:54
K.MK.M
693412
asked Nov 28 '18 at 5:54
K.MK.M
693412
asked Nov 28 '18 at 5:54
K.MK.M
693412
693412
2
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
1
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09
add a comment |
2
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
1
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09
2
2
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
1
1
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09
add a comment |
1 Answer
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The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $rsqrt{r} = r^{3/2}$, and it becomes clear.
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
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$begingroup$
The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $rsqrt{r} = r^{3/2}$, and it becomes clear.
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
$endgroup$
add a comment |
$begingroup$
The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $rsqrt{r} = r^{3/2}$, and it becomes clear.
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
$endgroup$
add a comment |
$begingroup$
The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $rsqrt{r} = r^{3/2}$, and it becomes clear.
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
$endgroup$
The sum of a finite geometric series with ratio $x$ is given by
$$1 + x + x^2 + ... + x^n = frac{x^{n+1} - 1}{x-1}$$
Notice that you have a geometric series, with ratio $rsqrt{r} = r^{3/2}$, and it becomes clear.
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
answered Nov 28 '18 at 5:56
Eevee TrainerEevee Trainer
5,7911936
5,7911936
add a comment |
add a comment |
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2
$begingroup$
Summation formula for a geometric series.
$endgroup$
– Thomas Shelby
Nov 28 '18 at 5:56
1
$begingroup$
@ThomasShelby: Thanks, I got confused because it looked different from the usual formula for a geometric series.
$endgroup$
– K.M
Nov 28 '18 at 6:09