# 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.

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

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

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

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

$\text{Step - 3 :}$ 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 $n$ terms. $\large (1 \cdot 2 \cdot 3 \cdot 4) + (2 \cdot 3 \cdot 4 \cdot 5) + (3 \cdot 4 \cdot 5 \cdot 6) + \quad ...$

Solution :

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

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

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

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

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

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

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

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

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

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. $\begin{array}{c}~ 5t_1 & = & {\color{#3D99F6}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} & - & {\color{#20A900}0} \\ 5t_2 & = & {\color{#E81990}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} & - & {\color{#3D99F6}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} \\ 5t_3 & = & {\color{#D61F06}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} & - & {\color{#E81990}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} \\ 5t_4 & = & {\color{#69047E}\cancel{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}} & - & {\color{#D61F06}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} \\ & & & \vdots \\ 5t_{n - 1} & = & {\color{#EC7300}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} & - & {\color{teal}\cancel{(n - 2) \cdot (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2)}} \\ 5t_n & = & {\color{#20A900}n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3) \cdot (n + 4)} & - & {\color{#EC7300}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 5({\color{#3D99F6}t_1 + t_2 + t_3 + ... + t_n}) & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ 5{\color{#3D99F6}S_n} & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ & & \boxed{\color{#20A900} S_n = \dfrac{n(n + 1)(n + 2)(n + 3)(n + 4)}{5}} \\ \end{array}$

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 !!!

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

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 : $\dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + ...$ WORKING RULE :

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

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

$\text{Step - 3 :}$ 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. $\dfrac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \dfrac{1}{2 \cdot 3 \cdot 4 \cdot 5} + \dfrac{1}{3 \cdot 4 \cdot 5 \cdot 6} + ...$

Solution :

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

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

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

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

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

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

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

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

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

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

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. $\begin{array}{c}~ 3t_1 & = & {\color{#20A900}\dfrac{1}{1 \cdot 2 \cdot 3}} & - & {\color{#3D99F6}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} \\ 3t_2 & = & {\color{#3D99F6}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} & - & {\color{#E81990}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} \\ 3t_3 & = & {\color{#E81990}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} & - & {\color{#D61F06}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} \\ 3t_4 & = & {\color{#D61F06}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} & - & {\color{#69047E}\cancel{\dfrac{1}{5 \cdot 6 \cdot 7}}} \\ &&& \vdots \\ 3t_{n - 1} & = & {\color{teal}\cancel{\dfrac{1}{(n - 1) \cdot n \cdot (n + 1)}}} & - & {\color{#EC7300}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} \\ 3t_n & = & {\color{#EC7300}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} & - & {\color{#20A900}\dfrac{1}{(n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 3({\color{#3D99F6}t_1 + t_2 + t_3 + ... + t_n}) & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ 3S_n & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ & & \boxed{\color{#20A900} S_n = \dfrac{1}{18} - \dfrac{1}{3(n + 1)(n + 2)(n + 3)}} \\ \end{array}$

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 $n$ but you can find the sum by taking number of terms as $n$ and at the ant substitute the value of $n$ 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

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

Note by Ram Mohith
2 years ago

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

• Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
• Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
• Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
• Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$ ... $$ or $ ... $ to ensure proper formatting.
2 \times 3 $2 \times 3$
2^{34} $2^{34}$
a_{i-1} $a_{i-1}$
\frac{2}{3} $\frac{2}{3}$
\sqrt{2} $\sqrt{2}$
\sum_{i=1}^3 $\sum_{i=1}^3$
\sin \theta $\sin \theta$
\boxed{123} $\boxed{123}$

Sort by:

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

Dear Farhat.

- 1 year, 11 months ago

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.

- 1 year, 11 months ago

What is a contributions page?

- 1 year, 11 months ago

Awesomely wrote with Latex. Keep it up

- 1 year, 12 months ago

@Calvin Lin, where is the save option?

- 2 years ago

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.

Staff - 2 years ago

@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

- 2 years ago

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

$1^3 + 2^3 + 3^3 \cdots n^3 = (1 + 2 + 3 \cdots n)^2$

You do not really have to memorise the formula

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Yes. Thank you very much.

- 2 years ago

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

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

- 2 years ago

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.

I second you!

- 2 years ago

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.

- 2 years ago

Math feeds us but sometimes it is out of our grasp

- 2 years ago

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

- 2 years ago

Math knowledge is indeed fueled by passion

- 2 years ago

Very correct, Farhat. Intelligent oration there.

- 2 years ago

My thanks to you

- 2 years ago

No problem, friend.

- 2 years ago

MATH - Means, Always, The Heaven

- 2 years ago

True but not more than science for me.

- 2 years ago

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

- 2 years ago

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 !!!

- 2 years ago

Exceptionally excellent. Should be a Master peice.

- 2 years ago

Thank you, brother.

- 2 years ago

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

- 2 years ago

Two young dynamites.

- 2 years ago

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

- 2 years ago

WOW! You have great poetry skills

- 2 years ago

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

- 2 years ago

Ok. But I don't understand the rules.

- 2 years ago

Which part of the rules?

- 2 years ago

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

- 2 years ago

Oh! The structure is like [link text ]then you add (link) ($[link text](link)$)

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

- 2 years ago

Ok. Something like EKENE FRANKLIN?

- 2 years ago

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.

- 2 years ago

Let me try again. @Ram Mohith.

- 2 years ago

It is not still working.

- 2 years ago

The link is the URL you see above this website

- 2 years ago

You mean brilliant.org?

- 2 years ago

Tap on that link

- 2 years ago

It is taking me to the home page!

- 2 years ago

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

- 2 years ago

Ok.

- 2 years ago

Now, do you understand?

- 2 years ago

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

- 2 years ago

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

- 2 years ago

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).

People who are like me! (I am 10)

- 2 years ago

I like this note!

- 2 years ago

Thank You

- 2 years ago

I made an inspirational note about the Collatz Conjecture

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Yes I have seen it. It is a good one

- 2 years ago

How was the vacation? Did it go well?

- 2 years ago

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

- 2 years ago

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

- 2 years ago

Open problem group. I have joined it already.

- 2 years ago

Which group? Can you share it's link?

- 1 year, 7 months ago

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

- 1 year, 7 months ago

Also, what is contribution page and how is it made?

- 1 year, 7 months ago

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

- 1 year, 7 months ago

Ok thank you!

- 1 year, 7 months ago

Ok!

- 2 years ago

My thanks to you.

- 2 years ago

×