Asked by Elosy
Prove the following by mathematical induction for all integers (n>or=1)
1^3+2^3+3^3+-----+n^3=(n(n+1)/2)^2
1^3+2^3+3^3+-----+n^3=(n(n+1)/2)^2
Answers
Answered by
oobleck
is it true for k=1?
1^3 = ((1*2)/2)^2 ✅
So, assuming true for k=n, what about k=n+1?
1^3 + ... + n^3 + (n+1)^3 = (n(n+1)/2)^2 + (n+1)^3
= 1/4 n^2 (n+1)^2 + (n+1)^3
= 1/4 (n^4 + 2n^3 + n^2 + n^3 + 3n^2 + 3n + 1)
= 1/4 (n^4 + 3n^3 + 4n^2 + 3n + 1)
= 1/4 (n+1)^2 (n+2)^2
= ((n+1)(n+1+1)/2)^2
= (k(k+1)/2)^2
1^3 = ((1*2)/2)^2 ✅
So, assuming true for k=n, what about k=n+1?
1^3 + ... + n^3 + (n+1)^3 = (n(n+1)/2)^2 + (n+1)^3
= 1/4 n^2 (n+1)^2 + (n+1)^3
= 1/4 (n^4 + 2n^3 + n^2 + n^3 + 3n^2 + 3n + 1)
= 1/4 (n^4 + 3n^3 + 4n^2 + 3n + 1)
= 1/4 (n+1)^2 (n+2)^2
= ((n+1)(n+1+1)/2)^2
= (k(k+1)/2)^2
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