Welcome to my article **How to solve 1^3 + 2^3 + 3^3+4^3+5^3 to n^3 formula calculator**. This question is taken from the simplification lesson.

The solution of this question 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 + 3^3+4^3+5^3 to n^3 formula calculator**, read and understand it carefully till the end.

Let us know how to solve this question **How to solve 1^3 + 2^3 + 3^3+4^3+5^3 to n^3 formula calculator**.

First write the question on the page of the notebook.

## How to solve 1^3 + 2^3 + 3^3+4^3+5^3 to n^3 formula calculator

**To solve this question, first we will write in this way**,

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

\displaystyle {{1}^{3}}\text{ }=\text{ }11^3+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**

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 \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.**

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