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,
\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{ }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
\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 \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.
https://dhams.in/how-to-solve-122232-to-n2-formula/
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