Can anyone prove this using mathematical induction 1^3+2^3+…+n^3 = n(n+1)……..^2?

Question

Can anyone prove this using mathematical induction:
1^3+2^3+…+n^3 = n(n+1)……..^2
……………………---------
………………………..2
for all integers n >=1.

Note n(n+1) / 2 are in brackets and than all that is squared by 2 (^2).

Answer 1

To prove this by mathematical induction, we need two things:
a) Show it is true when n = 1
b) Show that if it is true for n, then it is true for n+1

So, when n = 1,
1^3 = 1
and (1*2/2)^2 = (2/2)^2 = 1^2 = 1
So 1^3 = (1*2/2)^2

Now the hard part.
Assume that 1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2
Then 1^3 + 2^3 + ... + n^3 + (n+1)^3 = [n(n+1)/2]^2 + (n+1)^3
= [n^2 + n]^2/2^2 + [n^3 + 3n^2 + 3n + 1]
= [n^4 + 2n^3 + n^2]/4 + 4[n^3 + 3n^2 + 3n + 1]/4
= [n^4 + 6n^3 + 13n^2 + 12n + 4]/4
= [n^2 + 3n +2]^2/(2^2)
= { [n^2 + 3n +2]/2 }^2
= { (n+1)(n+2)/2 }^2

So 1^3 + 2^3 + ... + n^3 + (n+1)^3 = { (n+1)(n+2)/2 }^2
Therefore, we have shown both parts a) and b), so
1^3 + 2^3 + ... + n^3 = [n(n+1)/2]^2 is true for all integers n such that n >= 1.

I hope this helps!

Answer 2

This statement seems flawed; consider n = 3 1^3 + 3^3 = 1 + 27 = 28 but (3(3+1) / 2)^2 = 6^2 = 36. Are you sure this is the right problem? Edit: Above poster seems to have identified what you may have meant your question to be. Looks good.

Answer 3

P(n): 1³ + 2³ + 3³-------n³ = [ n(n + 1) / 2 ] ²
Let k be a value of n:-
P(k): 1³ + 2³ +3³ -----k³ = [ k.(k + 1) / 2 ] ²
Have now to show P(1) true and P(k + 1) true.

Consider P(1)
LHS = 1³ = 1
RHS = [1 x 2 /2 ]² = 1
Thus LHS = RHS

Consider P(k + 1)
1³ + 2³ + 3³ ----(k + 1)³ = [(k + 1).(k + 2) / 2]²
This has now to be proven:-
LHS = 1³ + 2³ + --k³ + (k + 1)³
RHS = [ (k.(k + 1) / 2)² + (k + 1)³ ]
RHS = (k + 1)².[ k² / 4 + (k + 1) ]
RHS = (k + 1)².[ (k² + 4k + 4 ) / 4 ]
RHS = [(k + 1).(k + 2) / 2]²
Thus P(k + 1) is true

P(k) true, P(1) true, P(k + 1) true
Thus P(n) true.

Answer 4

a) the statement must be true for the integer 1:
1^3 = 1
(1(1+1)/2)^2 = 1

b) if the statement is true for integers 1,2...k, then it is true for k+1
S(k) = ((k(k+1))/2)^2
S(k+1) = S(k) + (k+1)^3
= ((k(k+1))/2)^2 + k^3 + 3k^2 + 3k + 1
= (k^4 + 2k^3 + k^2)/4 + k^3 + 3k^2 + 3k + 1
= (1/4)k^4 + (3/2)k^3 + (13/4)k^2 + 3k + 1
= (1/4)(k^4 + 6k^3 + 13k^2 + 12k + 4)
= (1/4)(k+1)^2(k+2)^2
= ((k+1)(k+2)/4)^2

This is the same as the formula (n(n+1)/2)^2, except k+1 is substituted in for n, thus the formula has been proven by mathematical induction.

Answer 5

Prove: 1^3+2^3+…+n^3 = [n(n+1)/2]^2
n = 1, LHS = 1 and RHS = 1. It works.
Assume at n = k, it works too.
1^3+2^3+…+k^3 = [k(k+1)/2]^2
At n = k+1,
1^3+2^3+…+k^3 + (k+1)^3
= [k(k+1)/2]^2+(k+1)^3
= (k+1)^2[k^2+4k+4]/2^2
= [(k+1)(k+2)/2]^2

End of proof.