Mathematical Induction Proof: P(n): (1^3)+(2^3)+...+(n^3)=(1+2+...+n)^2 for all n

Prove: P(n): (1^3)+(2^3)+...+(n^3)=(1+2+...+n)^2 for all natural numbers n

Attempt so far:

1. Basis Step - P(1) is true since 1^3=1^2

2. Induction Step - NTS P(k)--> P(k+1)

3. Suppose (1^3)+(2^3)+...+(k^3)=(1+2+...+k)^2 is true for some k, then we NTS P(k+1) is true.

4. P(k+1): (1^3)+(2^3)+...+(k^3)+((k+1)^3)=(1+2+...+(k+1))^2

5. From our supposition, the LHS becomes:

6. P(k+1): (1+2+...+k)^2 + (k+1)^3 = (1+2+...+(k+1))^2

This is where I'm stuck. Also, on step four, why does the LHS have two k's (the k and k+1), while the RHS only has one? There's a similar example that does this, but I don't know why I did it here.

Any help would be appreciated. Thanks! (Nod)

Re: Mathematical Induction Proof: P(n): (1^3)+(2^3)+...+(n^3)=(1+2+...+n)^2 for all n

Hi

First of all U should consider that :

1+2+3+. . . +n=n(n+1)/2 [1]

(can be simply proven by induction),

all we have to prove is:

1^3+2^3+...+k^3+(k+1)^3=(1+2+...+k+k+1)^2 [2]

According to hypothesis the LHS of the equation above can be written as:

(1+2+...+k)^2+(k+1)^3 ,

which referring to [1] can be reviewed as:

(k(k+1)/2)^2+(k+1)^3=((k+1)(k+2)/2)^2.

which is identical to RHS of [2] and proof is complete. :)