How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2

Our mission is to systematically share mathematics information to people around the world and to make it universally accessible and useful.
The solution of this question How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2 has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2 , read and understand it carefully till the end.

Let us know how to solve this question How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2 .

First write the question on the page of the notebook.

How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2

To solve this question, we will write this question in the simplest form as follows and simplify it.

\displaystyle {{1}^{3}}+{{2}^{3}}+{{3}^{3}}+{{4}^{3}}+{{5}^{3}}+………..+{{n}^{3}}

A remarkable discovery is suggested by examining the first five formulations:

\displaystyle {{1}^{3}}\text{ }=\text{ }1

1^3+2^3=9

1^3+2^3+3^3=36

1^3+2^3+3^3+4^3=100

1^3+2^3+3^3+4^3+5^3=225

It seems that the sum is always square, but what is even more remarkable is that the sum of the first n cubes, 1^3+2^3+…+n3 = \displaystyle {{\left( {\frac{{n(n+1)}}{2}} \right)}^{2}}, which is the nth The square of the triangle number.

For example,

1^3+2^3+…+10^3

= \displaystyle {{\left( {\frac{{10(10+1)}}{2}} \right)}^{2}}

= \displaystyle {{\left( {5\times 11} \right)}^{2}}

= \displaystyle {{\left( {55} \right)}^{2}}.

= 3025

We will prove this result deductively, using the same method used to prove that formula for the sum of squares; It is hoped that this will provide some insight into how else the chain of powers can be found.

Proof

\displaystyle \sum\limits_{{r=1}}^{n}{{{{r}^{4}}-(r-1)={{n}^{4}}-{{{(n-1)}}^{4}}+{{{(n-1)}}^{4}}-{{{(n-2)}}^{4}}+………+{{3}^{4}}-{{2}^{4}}+{{2}^{4}}-{{1}^{4}}+{{1}^{4}}-{{0}^{4}}={{n}^{4}}}}

But

\displaystyle {{r}^{4}}{{\left( {r1} \right)}^{4}}\text{ }=\text{ }{{r}^{4}}\text{ }\text{ }\left( {{{r}^{4}}4{{r}^{3}}+6{{r}^{2}}4r+1} \right)\text{ }=\text{ }4{{r}^{3}}6{{r}^{2}}+4r1.

\displaystyle \therefore ~\sum ~4{{r}^{3}}6{{r}^{2}}+4r1=4\sum ~{{r}^{3}}~\text{ }6\sum ~{{r}^{2}}~+\text{ }4\sum ~r~~\sum 1
See also  How to solve 1+2/9+1/7+x=0

\displaystyle ~\qquad =4\sum ~{{r}^{3}}~\text{ }6n(n+1)(2n+1)/6\text{ }+\text{ }4n(n+1)/2\text{ }n

\displaystyle ~\qquad =4\sum ~{{r}^{3}}~~n(n+1)(2n+1)\text{ }+\text{ }2n(n+1)\text{ }n

\displaystyle \qquad ={{n}^{4}}

\displaystyle \therefore \text{ }4\sum ~{{r}^{3}}~=~\qquad {{n}^{4}}~+~n(n+1)(2n+1)\text{ }\text{ }2n(n+1)\text{ }+n

\displaystyle n({{n}^{3}}+\text{ }(n+1)(2n+1)\text{ }\text{ }2(n+1)\text{ }+\text{ }1

\displaystyle =\qquad n({{n}^{3}}~+\text{ }2{{n}^{2}}+3n+1\text{ }\text{ }2n2\text{ }+\text{ }1)

\displaystyle =\qquad n({{n}^{3}}+2{{n}^{2}}+n)

\displaystyle =\qquad {{n}^{2}}({{n}^{2}}+2n+1)

\displaystyle =\qquad {{n}^{2}}{{(n+1)}^{2}}

so,

\displaystyle \therefore \sum {{r}^{3}}~=\frac{{{{n}^{2}}{{{(n+1)}}^{2}}}}{4}

\displaystyle ={{\left( {\frac{{(n(n+1)}}{2}} \right)}^{2}}

In other words, the sum of the first n cubes is the square of the sum of the first n natural numbers.

mathwaycalculus

This article How to solve 1^3+2^3+…+n^3= n(n+1)/2 ^2 has been completely solved by tireless effort from our side, still if any error remains in it then definitely write us your opinion in the comment box. If you like or understand the methods of solving all the questions in this article, then send it to your friends who are in need.

Note: If you have any such question, then definitely send it by writing in our comment box to get the answer.
Your question will be answered from our side.

Thank you once again from our side for reading or understanding this article completely.

Leave a Comment