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.

\(\Large \color{green} \mathcal{TYPE : 1}\)

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 \(\boxed{{\color{blue}\text{Next Factor of last term}} - \color{pink}\text{Previous Factor of first term}}\) .

\(\text{Step - 3 :}\) 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. \[\begin{array} ~\require{cancel} 5t_1 & = & {\color{blue}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} & - & {\color{green}0} \\ 5t_2 & = & {\color{pink}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} & - & {\color{blue}\cancel{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5}} \\ 5t_3 & = & {\color{red}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} & - & {\color{pink}\cancel{2 \cdot 3 \cdot 4 \cdot 5 \cdot 6}} \\ 5t_4 & = & {\color{purple}\cancel{4 \cdot 5 \cdot 6 \cdot 7 \cdot 8}} & - & {\color{red}\cancel{3 \cdot 4 \cdot 5 \cdot 6 \cdot 7}} \\ & & & \vdots \\ 5t_{n - 1} & = & {\color{orange}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} & - & {\color{teal}\cancel{(n - 2) \cdot (n - 1) \cdot n \cdot (n + 1) \cdot (n + 2)}} \\ 5t_n & = & {\color{green}n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3) \cdot (n + 4)} & - & {\color{orange}\cancel{(n - 1) \cdot n \cdot (n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 5({\color{blue}t_1 + t_2 + t_3 + ... + t_n}) & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ 5{\color{blue}S_n} & = & n(n + 1)(n + 2)(n + 3)(n + 4) \\ & & \boxed{\color{green} S_n = \dfrac{n(n + 1)(n + 2)(n + 3)(n + 4)}{5}} \\ \end{array}\]

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 !!!

\(\Large \color{green} \mathcal{TYPE : 2}\)

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 \(\boxed{{\color{blue}\text{Last Factor}} - \color{pink}\text{First Factor}}\) .

\(\text{Step - 3 :}\) 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. \[\begin{array} ~ \require{cancel} 3t_1 & = & {\color{green}\dfrac{1}{1 \cdot 2 \cdot 3}} & - & {\color{blue}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} \\ 3t_2 & = & {\color{blue}\cancel{\dfrac{1}{2 \cdot 3 \cdot 4}}} & - & {\color{pink}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} \\ 3t_3 & = & {\color{pink}\cancel{\dfrac{1}{3 \cdot 4 \cdot 5}}} & - & {\color{red}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} \\ 3t_4 & = & {\color{red}\cancel{\dfrac{1}{4 \cdot 5 \cdot 6}}} & - & {\color{purple}\cancel{\dfrac{1}{5 \cdot 6 \cdot 7}}} \\ &&& \vdots \\ 3t_{n - 1} & = & {\color{teal}\cancel{\dfrac{1}{(n - 1) \cdot n \cdot (n + 1)}}} & - & {\color{orange}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} \\ 3t_n & = & {\color{orange}\cancel{\dfrac{1}{n \cdot (n + 1) \cdot (n + 2)}}} & - & {\color{green}\dfrac{1}{(n + 1) \cdot (n + 2) \cdot (n + 3)}} \\ \hline \\ 3({\color{blue}t_1 + t_2 + t_3 + ... + t_n}) & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ 3S_n & = & \dfrac{1}{6} & - & \dfrac{1}{(n + 1)(n + 2)(n + 3)} \\ & & \boxed{\color{green} S_n = \dfrac{1}{18} - \dfrac{1}{3(n + 1)(n + 2)(n + 3)}} \\ \end{array}\]

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

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## Comments

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TopNewest@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

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

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I like this note!

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Thank You

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I made an inspirational note about the Collatz Conjecture

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

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

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@Ram Mohith,

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

Dear Farhat.

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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.

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What is a contributions page?

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Awesomely wrote with Latex. Keep it up

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@Calvin Lin, where is the save option?

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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.

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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

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

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Yes. Thank you very much.

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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.

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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.

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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.

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I second you!

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

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

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EKENE FRANKLIN?

Ok. Something likeLog in to reply

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@Ram Mohith.

Let me try again.Log in to reply

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@EKENE FRANKLIN!

SoLog in to reply

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

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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).

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People who are like me! (I am 10)

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