Find the sum or value of following expression [closed]
$begingroup$
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
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
$endgroup$
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.
add a comment |
$begingroup$
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
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
$endgroup$
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
add a comment |
$begingroup$
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
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
$endgroup$
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
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
edited Nov 25 '18 at 4:39
Rócherz
2,9112821
2,9112821
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
asked Nov 25 '18 at 4:27
Pravin KumarPravin Kumar
474
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $sqrt{2}+1$.
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
$endgroup$
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $sqrt{2}+1$.
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
$endgroup$
add a comment |
$begingroup$
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 $sqrt{2}+1$.
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
$endgroup$
add a comment |
$begingroup$
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 $sqrt{2}+1$.
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
$endgroup$
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 $sqrt{2}+1$.
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
edited Nov 25 '18 at 5:48
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
answered Nov 25 '18 at 5:14
achille huiachille hui
96.2k5132260
96.2k5132260
add a comment |
add a comment |
$begingroup$
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