How do you proof that (1^3+2^3+3^3…n^3) equals to (1+2+3+…n) ^2?

Answer 1

Proof by Induction

First,

1^3 = 1 = 1^2

1^3 + 2^3 = 1 + 8 = 9 = 3^2 = (1+2)^2

1^3+ 2^3 + 3^3 = 1+8+27 = 36 = 6^2 = (1+2+3)^2

So the statement is true for n=1,2,3

GIVEN: (1^3+2^3+ .... + k^3) = (1+2+3+...+k)^2 is the induction hypothesis...

Recall the formula for the sum of the first n integers is 1+2+3+4+...+n = n(n+1)/2

Then

{(1^3+2^3+ .... + k^3} + (k+1)^3 = (1+2+3+...+k)^2 + (k+1)^3 <--- substitution

= [k(k+1)/2]^2 + (k+1)^3 <--- sum of the first k integers inside parenthesis

= k^2 (k+1)^2 /4 + (k+1)^3 <--- squares first term

= {k^2 (k+1)^2 + 4(k+1)^3} / 4 <--- common denominator is 4

= (k+1)^2 (k^2 + 4k + 4) / 4 <--- factors out (k+1)^2 in numerator

= (k+1)^2 (k+2)^2/ 4 <--- factors 2nd polynomial in numerator

= [(k+1)(k+2)/2 ]^2 <--- law of exponents

= [ 1 + 2 + 3 + .... + k+1 ] ^2 <--- sum of first k+1 integers