How to Solve Complicated Summations ?

As you all know the of sum of first \(n\) natural numbers is given by the formula \(\dfrac{n(n + 1)}{2}\), the sum of their squares is given by the formula \(\dfrac{n(n + 1)(2n + 1)}{6}\), the sum of their cubes is given by the formula \(\left ( \dfrac{n(n + 1)}{2} \right )^2\). Using these three formulas we used to solve questions based on summation easily. But what if the expression which has to be summed has higher orders like \(n^4, n^5, etc\). You can derive the sum of \(4^{th}\) powers with some difficulty but it will be impossible to do it for fifth powers and so on. So, in that situations we are not going to solve the series using formulae rather we go by technique. The main concept we should remember while doing by this technique is that we should express every term of the given series as difference of two factors so that all terms will get cancelled except the first one and last one. There are two types of methods based on the given series. We will discuss about them more clearly here.


TYPE:1\Large \color{#20A900} \mathcal{TYPE : 1}

The main idea in this method is, splitting the nthn^{th} term as a difference of two terms. This method can be applied to a series in which each term is composed of rr factors which are in AP, the first factor of the several terms being in AP. An example for such series is given below : (1×2×3×4)+(2×3×4×5)+...(1 \times 2 \times 3 \times 4) + (2 \times 3 \times 4 \times 5) + ... WORKING RULE :

Step - 1 :\text{Step - 1 :} Write down the nthn^{th} term

Step - 2 :\text{Step - 2 :} Multiply both the sides with Next Factor of last termPrevious Factor of first term\boxed{{\color{#3D99F6}\text{Next Factor of last term}} - \color{#E81990}\text{Previous Factor of first term}} .

Step - 3 :\text{Step - 3 :} Split the RHS and arrange them in such a way that all the diagonal elements gets cancelled. And there you have got the answer.

Here is an illustration to explain this more clearly.

Illustration - 1 : Find the sum of the given below series up to nn terms. (1234)+(2345)+(3456)+...\large (1 \cdot 2 \cdot 3 \cdot 4) + (2 \cdot 3 \cdot 4 \cdot 5) + (3 \cdot 4 \cdot 5 \cdot 6) + \quad ...


Solution :

Our first job is find the nthn^{th} term. Here it is :

tn=n(n+1)(n+2)(n+3)t_n = n(n + 1)(n + 2)(n + 3)

Now, we we should multiply both the sides with Next Factor of last term - Previous Factor of first term    (n+4)Next factor of (n + 3)(n1)Previous factor of n\text{Next Factor of last term - Previous Factor of first term} \implies \underbrace{(n + 4)}_{\text{Next factor of (n + 3)}} - \underbrace{(n - 1)}_{\text{Previous factor of n}}

[(n+4)(n1)]tn=n(n+1)(n+2)(n+3)[(n+4)(n1)]5tn=n(n+1)(n+2)(n+3)(n+4)(n1)n(n+1)(n+2)(n+3)[(n + 4) - (n -1)]t_n = n(n + 1)(n + 2)(n + 3)[(n + 4) - (n - 1)] \\ 5t_n = n(n + 1)(n + 2)(n + 3)(n + 4) - (n - 1)n(n + 1)(n + 2)(n + 3)

Now put n=1n = 1, we will get 5t1=1234505t_1 = 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 - 0

Now put n=2n = 2, we will get 5t2=23456123455t_2 = 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6 - 1 \cdot 2 \cdot 3 \cdot 4 \cdot 5

Now put n=3n = 3, we will get 5t3=34567234565t_3 = 3 \cdot 4 \cdot 5 \cdot 6 \cdot 7 - 2 \cdot 3 \cdot 4 \cdot 5 \cdot 6

And so on, now put n=(n1)n = (n - 1), we will get 5tn1=(n1)n(n+1)(n+2)(n+3)(n2)(n1)n(n+1)(n+2)5t_{n - 1} = (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3) - (n - 2) \cdot (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2)

And now put n=nn = n, we will get 5tn=n(n+1)(n+2)(n+3)(n+4)(n1)n(n+1)(n+2)(n+3)5t_n = n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3) \cdot (n + 4) - (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)

Now, arrange each term in a separate line and see the magic !!! All the elements diagonally except first one and last will get cancelled while adding all the terms.  5t1=1234505t2=23456123455t3=34567234565t4=45678345675tn1=(n1)n(n+1)(n+2)(n+3)(n2)(n1)n(n+1)(n+2)5tn=n(n+1)(n+2)(n+3)(n+4)(n1)n(n+1)(n+2)(n+3)5(t1+t2+t3+...+tn)=n(n+1)(n+2)(n+3)(n+4)5Sn=n(n+1)(n+2)(n+3)(n+4)Sn=n(n+1)(n+2)(n+3)(n+4)5\begin{array}{c}~ 5t_1 & = & {\color{#3D99F6}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} & - & {\color{#20A900}0} \\ 5t_2 & = & {\color{#E81990}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} & - & {\color{#3D99F6}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} \\ 5t_3 & = & {\color{#D61F06}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} & - & {\color{#E81990}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} \\ 5t_4 & = & {\color{#69047E}\cancel{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}} & - & {\color{#D61F06}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} \\ & & & \vdots \\ 5t_{n - 1} & = & {\color{#EC7300}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} & - & {\color{teal}\cancel{(n - 2) \cdot (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2)}} \\ 5t_n & = & {\color{#20A900}n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3) \cdot (n + 4)} & - & {\color{#EC7300}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 5({\color{#3D99F6}t_1 + t_2 + t_3 + ... + t_n}) & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ 5{\color{#3D99F6}S_n} & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ & & \boxed{\color{#20A900} S_n = \dfrac{n(n + 1)(n + 2)(n + 3)(n + 4)}{5}} \\ \end{array}

You can see how every term with similar colors (except green) will get cancelled throughout leaving us only with two terms left.

This is how you have to solve these types of problems. Hope you have felt this quite good enough so let us move to the next set of problems !!!


TYPE:2\Large \color{#20A900} \mathcal{TYPE : 2}

In this type of series too the main idea is to split every term into factors so that every term gets cancelled. The difference between the before one and this one is that the terms will be in reciprocal. This method can be applied to a series in which each term is composed of the reciprocal of the product of 'r factors in AP, the first factor of the several terms being in the same AP. Here is an example : 11234+12345+13456+...\dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + ... WORKING RULE :

Step - 1 :\text{Step - 1 :} Write down the nthn^{th} term

Step - 2 :\text{Step - 2 :} Multiply both the sides with Last FactorFirst Factor\boxed{{\color{#3D99F6}\text{Last Factor}} - \color{#E81990}\text{First Factor}} .

Step - 3 :\text{Step - 3 :} Split the RHS and arrange them in such a way that all the diagonal elements gets cancelled. And there you have got the answer.

Here is an illustration to explain this more clearly.

Illustration - 2 : Find the sum of given below series up to n terms. 11234+12345+13456+...\dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + ...


Solution :

Our first job is find the nthn^{th} term. Here it is :

tn=1n(n+1)(n+2)(n+3)t_n = \dfrac{1}{n(n + 1)(n + 2)(n + 3)}

Now, we we should multiply both the sides with Last Factor - First Factor    (n+3)Last factornFirst factor\text{Last Factor - First Factor} \implies \underbrace{(n + 3)}_{\text{Last factor}} - \underbrace{n}_{\text{First factor}}

[(n+3)n]tn=(n+3)nn(n+1)(n+2)(n+3)    3tn=n+3n(n+1)(n+2)(n+3)nn(n+1)(n+2)(n+3)[(n + 3) - n]t_n = \dfrac{(n + 3) - n}{n(n + 1)(n + 2)(n + 3)} \implies 3t_n = \dfrac{n + 3}{n(n + 1)(n + 2)(n + 3)} - \dfrac{n}{n(n + 1)(n + 2)(n + 3)}

    3tn=1n(n+1)(n+2)1(n+1)(n+2)(n+3)\implies 3t_n = \dfrac{1}{n(n + 1)(n + 2)} - \dfrac{1}{(n + 1)(n + 2)(n + 3)}

Now put n=1n = 1, we will get 3t1=112312343t_1 = \dfrac{1}{1 \cdot 2 \cdot 3} - \dfrac{1}{2 \cdot 3 \cdot 4}

Now put n=2n = 2, we will get 3t2=123413453t_2 = \dfrac{1}{2 \cdot 3 \cdot 4} - \dfrac{1}{3 \cdot 4 \cdot 5}

Now put n=3n = 3, we will get 3t3=134514563t_3 = \dfrac{1}{3 \cdot 4 \cdot 5} - \dfrac{1}{4 \cdot 5 \cdot 6}

And so on now put, n=(n1)n = (n - 1), we will get 3tn1=1(n1)n(n+1)1n(n+1)(n+2)3t_{n - 1} = \dfrac{1}{(n - 1) \cdot n \cdot (n + 1)} - \dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}

And now put, n=nn = n, we will get 3tn=1n(n+1)(n+2)1(n+1)(n+2)(n+3)3t_n = \dfrac{1}{n \cdot (n + 1) \cdot (n + 2)} - \dfrac{1}{(n + 1) \cdot (n + 2) \cdot (n + 3)}

Now, arrange each term in a separate line and see the magic !!! All the elements diagonally except first one and last will get cancelled while adding all the terms.  3t1=112312343t2=123413453t3=134514563t4=145615673tn1=1(n1)n(n+1)1n(n+1)(n+2)3tn=1n(n+1)(n+2)1(n+1)(n+2)(n+3)3(t1+t2+t3+...+tn)=161(n+1)(n+2)(n+3)3Sn=161(n+1)(n+2)(n+3)Sn=11813(n+1)(n+2)(n+3)\begin{array}{c}~ 3t_1 & = & {\color{#20A900}\dfrac{1}{1 \cdot 2 \cdot 3}} & - & {\color{#3D99F6}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} \\ 3t_2 & = & {\color{#3D99F6}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} & - & {\color{#E81990}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} \\ 3t_3 & = & {\color{#E81990}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} & - & {\color{#D61F06}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} \\ 3t_4 & = & {\color{#D61F06}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} & - & {\color{#69047E}\cancel{\dfrac{1}{5 \cdot 6 \cdot 7}}} \\ &&& \vdots \\ 3t_{n - 1} & = & {\color{teal}\cancel{\dfrac{1}{(n - 1) \cdot n \cdot (n + 1)}}} & - & {\color{#EC7300}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} \\ 3t_n & = & {\color{#EC7300}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} & - & {\color{#20A900}\dfrac{1}{(n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 3({\color{#3D99F6}t_1 + t_2 + t_3 + ... + t_n}) & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ 3S_n & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ & & \boxed{\color{#20A900} S_n = \dfrac{1}{18} - \dfrac{1}{3(n + 1)(n + 2)(n + 3)}} \\ \end{array}

You can see how every term with similar colors (except green) will get cancelled throughout leaving us only with two terms left. In general I solved the problems taking the number of terms as nn but you can find the sum by taking number of terms as nn and at the ant substitute the value of nn as according to the question.


This is how we have to solve these types of series so now it's your turn to practice them. If you have any doubts you can keep them in the comments section below.

Want to improve in Mathematics !!! Want to see more of these !!! Then try my set Mathematics Done Right

Thank You !!!\Large \color{teal} \mathcal{Thank \ You \ !!!}

Note by Ram Mohith
1 year, 8 months ago

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I like this note!

Mohammad Farhat - 1 year, 8 months ago

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You know I really worked hard on writing such a complicated latex code !!!

Ram Mohith - 1 year, 8 months ago

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I have experienced this. See my note:π\pi, a beautiful number

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Yes I have seen it. It is a good one

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith My thanks to you.

Mohammad Farhat - 1 year, 8 months ago

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@Ram Mohith How was the vacation? Did it go well?

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Yes good. I planned to post this note before my vacation but it was not completed by then. So, today I finished it off.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith Did you join, Jason Dyers group? It is very useful (if you use it well)

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Open problem group. I have joined it already.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith Ok!

Mohammad Farhat - 1 year, 8 months ago

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@Ram Mohith Which group? Can you share it's link?

Mr. India - 1 year, 4 months ago

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@Mr. India Yaah sure I will share. Here it is : Open Problem Group

Ram Mohith - 1 year, 4 months ago

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@Ram Mohith Also, what is contribution page and how is it made?

Mr. India - 1 year, 4 months ago

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@Mr. India Contribution page tells you and other users who visit it the contributions you have done to brilliant or in other words your status in brilliant. It will give set about how many upvotes you have, how many people solved your questions, how many people have seen your notes and finally helpful reports, featured solutions, total comment upvotes.

To get contribution page you should have a minimum community contribution which I too don't know. You should either more upvotes or more solvers or both. (I got my contribution page when I have around 150 upvotes and 500 solvers to my questions

Ram Mohith - 1 year, 4 months ago

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@Ram Mohith Ok thank you!

Mr. India - 1 year, 4 months ago

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Thank You

Ram Mohith - 1 year, 8 months ago

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I made an inspirational note about the Collatz Conjecture

Mohammad Farhat - 1 year, 8 months ago

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@Ram Mohith, For the sum of cubes I think that

13+23+33n3=(1+2+3n)21^3 + 2^3 + 3^3 \cdots n^3 = (1 + 2 + 3 \cdots n)^2

You do not really have to memorise the formula

Mohammad Farhat - 1 year, 8 months ago

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Yes you are correct. x=1nx3=(x=1nx)2\displaystyle \sum_{x = 1} ^n x^3 = \left ( \displaystyle \sum_{x = 1} ^n x \right )^2

Ram Mohith - 1 year, 8 months ago

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Please have a look at Faulhaber polynomials. Finding general expressions for sums of kthk^{\text{th}} powers is not particularly difficult if you are familiar with certain properties of it, as well as the property of Bernoulli numbers.

A Former Brilliant Member - 1 year, 8 months ago

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I have posted this for those who deosn't know those things.

Ram Mohith - 1 year, 8 months ago

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People who are like me! (I am 10)

Mohammad Farhat - 1 year, 8 months ago

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Still, nothing wrong with alternate perspectives... you should always be open to other ideas, even from the most advanced of sources.

There is a relation between these sums and the inverse of a triangular matrix related to Pascal's triangle. I find this method far more accessible to people, as opposed to your approach, since the required tools are easily known to enthusiastic students; moreover, it captures the main essence of these sums, as studied by Faulhaber and Bernoulli himself. Furthermore, Conway has done a lot of work in this area, where he is able to relate the Derivative/Integral linear mapping of these sums with lower/higher-degree ones of the same form; this is quite a recent result from a paper of his in 2016, which I assume has been published (I do not recall the source directly, unfortunately).

A Former Brilliant Member - 1 year, 8 months ago

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I agree with Ram Mohith. Not everyone has a rapport with Faulhaber's polynomials. Besides, Ram has done a great work on the LaTeX.

EKENE FRANKLIN - 1 year, 8 months ago

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I second you!

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Yes, Farhat. For example, I am twelve, and have a weak ability in Faulhaber's formula's. Besides, Nigerians don't have a unique passion for math.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Math feeds us but sometimes it is out of our grasp

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat That is a true saying, Farhat, and is worthy of accolade.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Math knowledge is indeed fueled by passion

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Very correct, Farhat. Intelligent oration there.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin My thanks to you

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat No problem, friend.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin MATH - Means, Always, The Heaven

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat True but not more than science for me.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith Oh yeah! Science is the curiosity and wonders that is the size of whole worlds.

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Math and science are the heartbeat of the universe. They oxidise our world to a better place, reducing ignorance in the process. They catalyse the increase of knowledge. They adorn our lives to make them frictionless. Oh, what is better than math and science !!!

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin I think you make a great poet for my math poetry contest

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Ok. But I don't understand the rules.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Which part of the rules?

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat The truth is that I don't know how to create a link.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Oh! The structure is like [link text ]then you add (link) ([linktext](link)[link text](link))

Note: I added the latex code to prevent it from rendering but do not add the latex code

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Ok. Something like EKENE FRANKLIN?

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin I think some mistake happened. When I clicked it a page opens saying page not found. Check whether the link is correct or not. I prefer you to copy and paste the link rather than typing it so that there will be no errors.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith Let me try again. @Ram Mohith.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin It is not still working.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin The link is the URL you see above this website

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat You mean brilliant.org?

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Yes but you have to be very specific. (Brilliant.org is only part of the URL)

Mohammad Farhat - 1 year, 8 months ago

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@Ekene Franklin So @EKENE FRANKLIN!

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Tap on that link

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat It is taking me to the home page!

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Because you only included Brilliant.org. you have to copy the URL wholesale! (Copy - Paste would help)

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Ok.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Now, do you understand?

Mohammad Farhat - 1 year, 8 months ago

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@Ekene Franklin WOW! You have great poetry skills

Mohammad Farhat - 1 year, 8 months ago

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@Ekene Franklin Exceptionally excellent. Should be a Master peice.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith Thank you, brother.

EKENE FRANKLIN - 1 year, 8 months ago

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@Ekene Franklin Humanity is full of brothers and sisters. Not only connected by blood but also heart

Mohammad Farhat - 1 year, 8 months ago

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@Mohammad Farhat Two young dynamites.

Ram Mohith - 1 year, 8 months ago

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@Ram Mohith @Ram Mohith, Did you know that dynamites mean bombs (a type of bomb) that were discovered by Alfred Nobel (Nobel Prize creator)

Mohammad Farhat - 1 year, 8 months ago

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Please see my reply somewhere in this thread. Given anyone can click a few buttons, I think repeating what I have said here would be a pointless exercise.

A Former Brilliant Member - 1 year, 8 months ago

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Very great note, brother. Very informative to those ignorant of this.

EKENE FRANKLIN - 1 year, 8 months ago

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Yes. Thank you very much.

Ram Mohith - 1 year, 8 months ago

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@Calvin Lin, where is the save option?

Mohammad Farhat - 1 year, 8 months ago

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Can you send me a screenshot of what you see? The save function is still there.
I've tested this on other accounts / browsers, and have been unable to reproduce your issue.

Calvin Lin Staff - 1 year, 8 months ago

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@Calvin Lin, when it is on display mode(Thread), there is no save function but when I refresh the page there is a save function

Mohammad Farhat - 1 year, 8 months ago

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Awesomely wrote with Latex. Keep it up

Md Zuhair - 1 year, 8 months ago

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@Ram Mohith,

I FINALLY GOT A CONTRIBUTIONS PAGE! I AM SO EXCITED!

Dear Farhat.

Mohammad Farhat - 1 year, 8 months ago

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Congratulations. And keep it up. Now itself you have got 496 comment upvotes which is 8 upvotes less than mine. Looks like you will cross my contributions page soon.

Ram Mohith - 1 year, 8 months ago

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What is a contributions page?

EKENE FRANKLIN - 1 year, 8 months ago

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