[tex]\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}[/tex]
[tex]\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2[/tex]
Let
[tex]\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2[/tex]
For n=1
[tex]\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2[/tex]
[tex]\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2[/tex]
[tex]\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2[/tex]
[tex]\\ \sf\longmapsto P(1)=(1)^2[/tex]
[tex]\\ \bf\longmapsto P(1)=1=1^3[/tex]
Let k be any positive integer.
[tex]\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2[/tex]
We have to prove that p(k+1) is true.
consider
[tex]\sf 1^3+2^3\dots k^3+(k+1)^3[/tex]
[tex]\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3[/tex]
[tex]\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3[/tex]
[tex]\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}[/tex]
[tex]\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}[/tex]
[tex]\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}[/tex]
[tex]\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}[/tex]
[tex]\\ \sf\longmapsto \left(\dfrac{(k+1)(k+1+1)}{2}\right)^2[/tex]
[tex]\\ \sf\longmapsto (1^3+2^3+3^3\dots k^3)+(k+1)^3[/tex]
Thus P(k+1) is true whenever P(k) is true.
Hence by the Principal of mathematical induction statement P(n) is true for [tex]\bf n\epsilon N.[/tex]
Note:-
We can solve without simplifying the Question .I did it for clear steps and understanding .
https://brainly.com/question/13253046?
https://brainly.com/question/13347635?
Answer:
Step-by-step explanation:
[tex]1^3=1=\dfrac{1^2*(1+1)^2}{4} =\dfrac{4}{4} =1\\\\\\\displaystyle \sum_{i=1}^{n+1}\ i^3=(\sum_{i=1}^{n}\ i^3) + (n+1)^3\\\\\\=\dfrac{n^2*(n+1)^2}{4} +(n+1)^3\\\\\\=(n+1)^2*(\frac{n^2}{4} +n+1)\\\\\\=(n+1)^2*\dfrac{n^2+4n+4}{4} \\\\\\=\dfrac{(n+1)^2*(n+2)^2}{4}[/tex]
