Find the sum or value of following expression [closed]












1












$begingroup$


In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?



$$frac{{displaystyle sum_{n=1}^{99}} sqrt{10+sqrt{n}}}{{displaystyle sum_{n=1}^{99}} sqrt{10-sqrt{n}}}$$






summation







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$endgroup$



closed as off-topic by TheSimpliFire, user21820, Holo, Did, José Carlos Santos Nov 25 '18 at 10:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user21820, Holo, Did, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
    $endgroup$
    – David Peterson
    Nov 25 '18 at 4:36


















1












$begingroup$


In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?



$$frac{{displaystyle sum_{n=1}^{99}} sqrt{10+sqrt{n}}}{{displaystyle sum_{n=1}^{99}} sqrt{10-sqrt{n}}}$$






summation







share|cite|improve this question











$endgroup$



closed as off-topic by TheSimpliFire, user21820, Holo, Did, José Carlos Santos Nov 25 '18 at 10:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user21820, Holo, Did, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.
















  • $begingroup$
    You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
    $endgroup$
    – David Peterson
    Nov 25 '18 at 4:36
















1












1








1


5



$begingroup$


In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?



$$frac{{displaystyle sum_{n=1}^{99}} sqrt{10+sqrt{n}}}{{displaystyle sum_{n=1}^{99}} sqrt{10-sqrt{n}}}$$






summation







share|cite|improve this question











$endgroup$




In the problem I am not able to derive difference term. I know for solving summation we have to make difference term.
How to proceed with this problem?



$$frac{{displaystyle sum_{n=1}^{99}} sqrt{10+sqrt{n}}}{{displaystyle sum_{n=1}^{99}} sqrt{10-sqrt{n}}}$$






summation




summation






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 '18 at 4:39









Rócherz

2,9112821




2,9112821










asked Nov 25 '18 at 4:27









Pravin KumarPravin Kumar

474




474




closed as off-topic by TheSimpliFire, user21820, Holo, Did, José Carlos Santos Nov 25 '18 at 10:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user21820, Holo, Did, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.







closed as off-topic by TheSimpliFire, user21820, Holo, Did, José Carlos Santos Nov 25 '18 at 10:44


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – TheSimpliFire, user21820, Holo, Did, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.












  • $begingroup$
    You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
    $endgroup$
    – David Peterson
    Nov 25 '18 at 4:36




















  • $begingroup$
    You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
    $endgroup$
    – David Peterson
    Nov 25 '18 at 4:36


















$begingroup$
You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
$endgroup$
– David Peterson
Nov 25 '18 at 4:36






$begingroup$
You can replace 10 (N) by 2, and 99 (N^2-1) by 3. The result is $sqrt{2}+1$ for both. Examine the simpler case first, maybe.
$endgroup$
– David Peterson
Nov 25 '18 at 4:36












1 Answer
1






active

oldest

votes


















6












$begingroup$

This can be done using an identity I learned from math.SE.




For any $a > 0$ and $0 le b le a^2$, we have
$$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$




To prove this identity, just take squares on both sides and use the fact



$$begin{align}
left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
&= 2left(a + sqrt{a^2-b}right)end{align}$$



Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us



$$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
= sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
= sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
This leads to
$$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
= frac{1}{sqrt{2}-1} = sqrt{2}+1
$$
Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.






share|cite|improve this answer











$endgroup$




















    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

    This can be done using an identity I learned from math.SE.




    For any $a > 0$ and $0 le b le a^2$, we have
    $$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$




    To prove this identity, just take squares on both sides and use the fact



    $$begin{align}
    left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
    &= 2left(a + sqrt{a^2-b}right)end{align}$$



    Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us



    $$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
    = sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
    = sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
    This leads to
    $$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
    = frac{1}{sqrt{2}-1} = sqrt{2}+1
    $$
    Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.






    share|cite|improve this answer











    $endgroup$


















      6












      $begingroup$

      This can be done using an identity I learned from math.SE.




      For any $a > 0$ and $0 le b le a^2$, we have
      $$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$




      To prove this identity, just take squares on both sides and use the fact



      $$begin{align}
      left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
      &= 2left(a + sqrt{a^2-b}right)end{align}$$



      Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us



      $$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
      = sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
      = sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
      This leads to
      $$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
      = frac{1}{sqrt{2}-1} = sqrt{2}+1
      $$
      Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.






      share|cite|improve this answer











      $endgroup$
















        6












        6








        6





        $begingroup$

        This can be done using an identity I learned from math.SE.




        For any $a > 0$ and $0 le b le a^2$, we have
        $$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$




        To prove this identity, just take squares on both sides and use the fact



        $$begin{align}
        left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
        &= 2left(a + sqrt{a^2-b}right)end{align}$$



        Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us



        $$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
        = sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
        = sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
        This leads to
        $$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
        = frac{1}{sqrt{2}-1} = sqrt{2}+1
        $$
        Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.






        share|cite|improve this answer











        $endgroup$



        This can be done using an identity I learned from math.SE.




        For any $a > 0$ and $0 le b le a^2$, we have
        $$sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}} = sqrt{2}sqrt{a + sqrt{a^2-b}}$$




        To prove this identity, just take squares on both sides and use the fact



        $$begin{align}
        left(sqrt{a+sqrt{b}}+sqrt{a-sqrt{b}}right)^2 &= left(a + sqrt{b}right) + left(a - sqrt{b}right) + 2sqrt{a^2-b}\
        &= 2left(a + sqrt{a^2-b}right)end{align}$$



        Let $a$ be any integer $> 1$ and $c = a^2-b$. When we sum $b$ from $1$ to $a^2-1$, above identity tell us



        $$sum_{b=1}^{a^2-1} sqrt{a + sqrt{b}} + sum_{b=1}^{a^2-1}sqrt{a - sqrt{b}}
        = sqrt{2}sum_{b=1}^{a^2-1}sqrt{a + sqrt{a^2-b}}
        = sqrt{2}sum_{c=1}^{a^2-1}sqrt{a + sqrt{c}}$$
        This leads to
        $$frac{sum_{b=1}^{a^2-1}sqrt{a+sqrt{b}}}{sum_{b=1}^{a^2-1}sqrt{a-sqrt{b}}}
        = frac{1}{sqrt{2}-1} = sqrt{2}+1
        $$
        Substitute $a$ by $10$, this reduces to the ratio at hand. i.e. The ratio we seek equals to $sqrt{2}+1$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 25 '18 at 5:48

























        answered Nov 25 '18 at 5:14









        achille huiachille hui

        96.2k5132260




        96.2k5132260















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