How do I get $sqrt{r}(r-1) frac{r^{3n/2}-1}{r^{3/2}-1}$?












0












$begingroup$


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







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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
















0












$begingroup$


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







share|cite|improve this question











$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question











$endgroup$




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






share|cite|improve this question















share|cite|improve this question













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edited Nov 28 '18 at 5:58









Henning Makholm

239k17304541




239k17304541










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














  • 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










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.






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    active

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    active

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    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 28 '18 at 5:56









        Eevee TrainerEevee Trainer

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