Question

prove by mathematical induction that 1³+2³+3³+.......+n³=(n (n+1)/2)²​

Answer 1

This is the prove...

I hope this will satisfy you........

Answer 2

Given:

1³+2³+3³+...+n³=(n (n+1)/2)²​

To Find:

Proven by mathematical induction

Solution:

First, we should know the process of mathematical induction, for an equation P(n)

• The equation should be true for n=1
• If the equation is true for n=k then it is also true for n=k+1

Then equation P(n) will be true for all natural numbers and the equation will be proven by mathematical induction

So, for n=1, we have

$1^3=[\frac{n(n+1)}{2}]^2 \\1=[\frac{1*2}{2}] ^2\\1=1$

hence it is true for n=1

Now for n=k, it is true,

$1^3+2^3+3^3+...+k^3=[\frac{k(k+1)}{2}]^2$

so it needs to be true for n=k+1 also, we have,

$=1^3+2^3+3^3+...+k^3+(k+1)^3\\=[\frac{k(k+1)}{2}]^2+(k+1)^3\\=(k+1)^2[\frac{k^2+4k+4}{4} ]\\=(k+1)^2[\frac{(k+1+1)^2}{4}\\=[\frac{(k+1)(k+1+1)}{2}]^2$

Hence it is also true for n=k+1, so p(n) is true for all natural numbers.

Hence, Proved by mathematical induction that 1³+2³+3³+...+n³=(n (n+1)/2)²​.