Question #349929

let p(n) 1^3+2^3+....+n^3=(n(n+1)/2)2 for positive integer n


1
Expert's answer
2022-06-13T17:59:05-0400

Let P(n)P(n) be the proposition that the sum of the first nn positive integers,


13+23+....+n3=(n(n+1)2)21^3+2^3+....+n^3=(\dfrac{n(n+1)}{2})^2

Basis Step

P(1)P(1) is true because 13=1=(1(1+1)2)2.1^3=1=(\dfrac{1(1+1)}{2})^2.

Inductive Step

For the inductive hypothesis we assume that P(k)P(k) holds for an arbitrary positive integer k.k. That is, we assume that


13+23+....+k3=(k(k+1)2)21^3+2^3+....+k^3=(\dfrac{k(k+1)}{2})^2

Under this assumption, it must be shown that P(k+1)P(k + 1) is true, namely, that


13+23+....+k3+(k+1)3=((k+1)(k+1+1)2)21^3+2^3+....+k^3+(k+1)^3=(\dfrac{(k+1)(k+1+1)}{2})^2

is also true.

When we add k+1k + 1 to both sides of the equation in P(k),P(k), we obtain


13+23+....+k3+(k+1)3=1^3+2^3+....+k^3+(k+1)^3=

=(k(k+1)2)2+(k+1)3=(\dfrac{k(k+1)}{2})^2+(k+1)^3

=(k+12)2(k2+4k+4)=(\dfrac{k+1}{2})^2(k^2+4k+4)

=(k+12)2(k+2)2=(\dfrac{k+1}{2})^2(k+2)^2

=((k+1)(k+1+1)2)2=(\dfrac{(k+1)(k+1+1)}{2})^2

This last equation shows that P(k+1)P(k + 1) is true under the assumption that P(k)P(k) is true. This completes the inductive step.

We have completed the basis step and the inductive step, so by mathematical induction we know that P(n)P(n) is true for all positive integers n.n. That is, we have proved that


13+23+....+n3=(n(n+1)2)21^3+2^3+....+n^3=(\dfrac{n(n+1)}{2})^2

for all positive integers n.n.


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