Deriving the Identity: (1+2+3+...+n)^2 = 1^3 + 2^3 + 3^3 +...+ n^3

[DEMJR]

Solved: Is this an induction problem?

$(1+2+3+ \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3 , n \ge 1$

Provide a derivation of the identity above.

I do not know how to begin this problem. I tried to use induction but did not succeed. Also, I honestly do not know what it means by provide a derivation of the identity. Please do not give me the answer, I just need a helping hand in getting started.

 

[A. Bahat]

Use the fact that the sum of the first n positive integers is n(n+1)/2; then use induction.

 

[DEMJR]

I need to show that $k^3 + \big( \frac{(k+1)(k+2)}{2} \big)^2 = (k + 1)^3$ or am I way off on my induction basics?

I am following the guide I found at wolfram here http://demonstrations.wolfram.com/ProofByInduction/

However, I cannot get the algebra to work out where f(n) + a_(n+1) = f(n+1). Thank you for the help A. Bahat.

I keep getting a polynomial with degree four and I have no way to factor it into a cube.

 

[A. Bahat]

I would prove that 1+2³+3³+...+k³=k²(k+1)²/4 implies 1+2³+3³+...+(k+1)³=(k+1)²(k+2)²/4.

 

[DEMJR]

I am struggling on how to manipulate $k^2(k+1)^2/4 + (k+1)^3$ to equal $(k+1)^2(k+2)^2/4$

If I get a common denominator I get I am struggling on how to manipulate $(k^2(k+1)^24 + 4(k+1)^3)/4$

However, I cannot find the route that leads to f(k+1).

 

[micromass]

I am struggling on how to manipulate $k^2(k+1)^2/4 + (k+1)^3$ to equal $(k+1)^2(k+2)^2/4$

If I get a common denominator I get I am struggling on how to manipulate $(k^2(k+1)^24 + 4(k+1)^3)/4$

However, I cannot find the route that leads to f(k+1).

Use that

$$\frac{k^2(k+1)^2 + 4(k+1)^3}{4}=\frac{(k+1)^2(k^2+4(k+1))}{4}$$

This is distributivity. I just did ab+ac=a(b+c).

 

[DEMJR]

Thank you so much for your help micro and Bahat. I was overlooking distributivity. Thank you and I have it solved now. Not sure how I go about marking a thread as solved. I will try and do it by editing the title of my main post.