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Clues on how to solve these types of problems within 2-3 minutes for competitive exams
How to solve these?Find the area of the surfaceProve $int_0^x frac{f(u)(x-u)^n}{n!}du=int_0^x ( int_0^{u_n}( dotsb ( int_0^{u_1}f(t),dt ) du_1 ) dotsb )du_n$ via IBPHow would I go about evaluating $int frac{x}{(9-8x^2)^3}dx$?Evaluating $intfrac1{sqrt{12x + 0.02x^2}},mathrm dx$How to solve problems of this type?Solving integral without simplifying equationEvaluating a double integral of a complicated rational functionIntegration of a function approximated by a nth order polynomialShow that $frac3{16}sum_{n=1}^{infty}frac{(-1)^n-8}{n^3}=-frac{105}{64}zeta(3)$
$begingroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but I'd be glad if I could get some clues on how to solve this problem.
integration derivatives definite-integrals
edited 4 hours ago
Paras Khosla
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
$endgroup$
add a comment |
$begingroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but I'd be glad if I could get some clues on how to solve this problem.
integration derivatives definite-integrals
edited 4 hours ago
Paras Khosla
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
$endgroup$
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
7 hours ago
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
7 hours ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
6 hours ago
2
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
5 hours ago
add a comment |
$begingroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but I'd be glad if I could get some clues on how to solve this problem.
integration derivatives definite-integrals
edited 4 hours ago
Paras Khosla
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
$endgroup$
$$int_0^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}frac1{x-k}right),dx$$
I've tried solving this problem but only thing that comes to my mind is the manual integration by multiplication of the expressions which will literally take much longer than the allotted time for competitive exams Now this is a homework and exercises problem but I'd be glad if I could get some clues on how to solve this problem.
integration derivatives definite-integrals
integration derivatives definite-integrals
edited 4 hours ago
Paras Khosla
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
edited 4 hours ago
Paras Khosla
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
edited 4 hours ago
Paras Khosla
1,344218
edited 4 hours ago
Paras Khosla
1,344218
edited 4 hours ago
Paras Khosla
1,344218
1,344218
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
asked 7 hours ago
HOME WORK AND EXERCISESHOME WORK AND EXERCISES
597
597
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
7 hours ago
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
7 hours ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
6 hours ago
2
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
5 hours ago
add a comment |
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
7 hours ago
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
7 hours ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
6 hours ago
2
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
5 hours ago
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
7 hours ago
$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
$endgroup$
– Lord Shark the Unknown
7 hours ago
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
7 hours ago
$begingroup$
The answer given is 101!-100! but no solutions also i can't find such problem online to learn
$endgroup$
– HOME WORK AND EXERCISES
7 hours ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
6 hours ago
$begingroup$
How about using the reverse product rule?
$endgroup$
– Paras Khosla
6 hours ago
2
2
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
5 hours ago
$begingroup$
As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
$endgroup$
– Paras Khosla
5 hours ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint:
By the product rule you have the following result:
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited 3 hours ago
Community♦
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
|
show 3 more comments
$begingroup$
Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
By the product rule you have the following result:
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited 3 hours ago
Community♦
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
|
show 3 more comments
$begingroup$
Hint:
By the product rule you have the following result:
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited 3 hours ago
Community♦
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
$endgroup$
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
|
show 3 more comments
$begingroup$
Hint:
By the product rule you have the following result:
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited 3 hours ago
Community♦
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
$endgroup$
Hint:
By the product rule you have the following result:
$$dfrac{mathrm d}{mathrm dx}prod_{k=1}^{100}(x-k)=left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)$$
Integrate both sides from $0$ to $102$, use the Fundamental Theorem of Calculus and you'll be done in no time.
edited 3 hours ago
Community♦
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
edited 3 hours ago
Community♦
1
edited 3 hours ago
Community♦
1
edited 3 hours ago
Community♦
1
1
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
answered 6 hours ago
Paras KhoslaParas Khosla
1,344218
1,344218
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
|
show 3 more comments
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
Thank you that was rather uncomplicated. :D
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
$begingroup$
I find the "both sides" confusing here. Both sides of what?
$endgroup$
– Henning Makholm
6 hours ago
1
1
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
$begingroup$
@HenningMakholm Integrate both sides of the following differential equation wrt $x$: $$int_{0}^{102}left(prod_{k=1}^{100}(x-k)right)left(sum_{k=1}^{100}dfrac{1}{(x-k)}right)mathrm dx=int_{0}^{102} mathrm d left(prod_{k=1}^{100}(x-k)right)$$
$endgroup$
– Paras Khosla
5 hours ago
3
3
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
$begingroup$
That's the most bizarrely convoluted way of saying "here is an antiderivative; use that that evaluate the definite integral" I've seen in a while. What on earth do you get out of describing it in that roundabout way?
$endgroup$
– Henning Makholm
3 hours ago
3
3
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
$begingroup$
@HenningMakholm Thinking this way helps when we encounter differential equations and you have to identify reverse chain rule, reverse product rule and a bunch of other "reverses". This essentially is another way of thinking about the same problem. "Convoluted way" is opinion-based and might not be for everyone.
$endgroup$
– Paras Khosla
3 hours ago
|
show 3 more comments
$begingroup$
Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
add a comment |
$begingroup$
Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
add a comment |
$begingroup$
Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Here's a quick hint: if you differentiate the product in the integrand, you get the entire integrand so by the fundamental theorem of calculus you can evaluate this very fast.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
answered 7 hours ago
Jonathan LevyJonathan Levy
1514
1514
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Jonathan Levy is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
add a comment |
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
$begingroup$
I think Paras said it--the product rule gives it to you.
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– Jonathan Levy
6 hours ago
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So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
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– HOME WORK AND EXERCISES
6 hours ago
$begingroup$
So how do i diffrentiate it? would'nt it take longer? I might sound stupid to you but I am new to these
$endgroup$
– HOME WORK AND EXERCISES
6 hours ago
2
2
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
$begingroup$
I think Paras said it--the product rule gives it to you.
$endgroup$
– Jonathan Levy
6 hours ago
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$begingroup$
My guess is the integrand is anti-symmetric about $x=51$ so that the integral is zero.
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– Lord Shark the Unknown
7 hours ago
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The answer given is 101!-100! but no solutions also i can't find such problem online to learn
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– HOME WORK AND EXERCISES
7 hours ago
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How about using the reverse product rule?
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– Paras Khosla
6 hours ago
2
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As a problem solving tip, always write out the first few terms of the series, make observations and hypothesize a probable path to the solution.
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– Paras Khosla
5 hours ago