Question

Prove that 1³+2³+....n³=n²(n+1)²/4. principle of mathematics induction​

Answer 1

The Simplified Question:-

$\sf 1^3+2^3\dots n^3=\dfrac{n^2(n+1)^2}{2}$

$\\ \sf\longmapsto 1^3+2^3+\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

Solution:-

Let

$\\ \sf\longmapsto P(n)=1^3+2^3\dots n^3=\left(\dfrac{n(n+1)}{2}\right)^2$

For n=1

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(1+1)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{1(2)}{2}\right)^2$

$\\ \sf\longmapsto P(1)=\left(\dfrac{2}{2}\right)^2$

$\\ \sf\longmapsto P(1)=(1)^2$

$\\ \bf\longmapsto P(1)=1=1^3$

Let k be any positive integer.

$\\ \sf\longmapsto P(k)= 1^3+2^3\dots k^3=\left(\dfrac{k(k+1)}{2}\right)^2$

We have to prove that p(k+1) is true.

consider

$\sf 1^3+2^3\dots k^3+(k+1)^3$

$\\ \sf\longmapsto \left(\dfrac{k(k+1)}{2}\right)^2+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2}{4}+(k+1)^3$

$\\ \sf\longmapsto \dfrac{k^2(k+1)^2+4(k+1)^3}{4}$

$\\ \sf\longmapsto \dfrac{k+1)^2\left\{k^2+4k+4\right\}}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+2)^2}{4}$

$\\ \sf\longmapsto \dfrac{(k+1)^2(k+1+1)^2}{4}$

$\\ \sf\longmapsto \left(\dfrac{(k+1)(k+1+1)}{2}\right)^2$

$\\ \sf\longmapsto (1^3+2^3+3^3\dots k^3)+(k+1)^3$

Thus P(k+1) is true whenever P(k) is true.

Hence by the Principal of mathematical induction statement P(n) is true for $\bf n\epsilon N.$

Note:-

We can solve without simplifying the Question .I did it for clear steps and understanding .

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Answer 2

Answer:

Step-by-step explanation:

$1^3=1=\dfrac{1^2*(1+1)^2}{4} =\dfrac{4}{4} =1\\\\\\\displaystyle \sum_{i=1}^{n+1}\ i^3=(\sum_{i=1}^{n}\ i^3) + (n+1)^3\\\\\\=\dfrac{n^2*(n+1)^2}{4} +(n+1)^3\\\\\\=(n+1)^2*(\frac{n^2}{4} +n+1)\\\\\\=(n+1)^2*\dfrac{n^2+4n+4}{4} \\\\\\=\dfrac{(n+1)^2*(n+2)^2}{4}$