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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.
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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





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,


= \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.


\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}}}}


\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}}


\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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