Prove Summation - Unknown factoring step (Prove that 1^3 + 2^3 + 3^3 + ... + n^3 = [n^2 ( n + 1 )^2] / 4)

[jddoxtator]

Here is the example question in full:

{Example 1: Prove Summation -

Prove that 1^3 + 2^3 + 3^3 + ... + n^3 = [n^2 ( n + 1 )^2] / 4

Step 1: When n = 1, the left side of the equation is 1^3 or 1.
The right side is [1^2 ( 1 + 1 )^2 ] / 4 or 1. Thus the statement is true for n = 1.

Step 2: Assume that 1^3 + 2^3 + 3^3 + ... + k^3 = [k^2 (k +1 )^2 ] / 4 for any natural number k.

Step 3: Show the given statement is true for n = k + 1.

1^3 + 2^3 + 3^3 + ... + k^3 = [k^2 ( k + 1 )^2 ] / 4 -Inductive hypothesis

1^3 + 2^3 + ... + k^3 + ( k + 1 )^3 = [k^2 ( k + 1 )^2 ] / 4 + ( k+1 )^3 -Add ( k + 1)^3 to each side
= [k^2 (k + 1)^2 + 4 ( k + 1 )^3 ] / 4 -The LCD is 4
= [ ( k + 1 )^2 [ k^2 + 4 ( k + 1 ) ] ] / 4 - Factor
= [ ( k + 1 )^2 ( k^2 + 4k + 4 ) ] / 4 -Simplify
= [ ( k + 1 )^2 ( k + 2 )^2 ] /4 -Factor
The last expression is the statement to be proved, where n has been replaced by k + 1. This proves the conjecture.}

Now, the failure in my understanding is the part in step three where we go from the step explained by -The LCD is 4 and the step explained by -Factor.

My issue is that I have no idea where the ^3 is going in the factoring step. To me it just looks like the commutative property performed on the first term and then k^2 is somehow grouped with 4 ( k + 1 ) while dropping the ^3.

Is there some short hand form of factoring I am not seeing here or is [ ( k + 1 )^3 ] just an identity of ( k + 1 )?

Very confused....

 

[jddoxtator]

Never mind.
After lots of reworking and watching a youtube video on this I figured it out.
It is the common factor of ( k + 1 )^2 being pulled from the whole expression in the numerator.
I feel dumb, well maybe not dumb, but mislead a bit. I feel like this was poorly explained by the example.

 

[khansaheb]

Never mind.
After lots of reworking and watching a youtube video on this I figured it out.
It is the common factor of ( k + 1 )^2 being pulled from the whole expression in the numerator.
I feel dumb, well maybe not dumb, but mislead a bit. I feel like this was poorly explained by the example.

While doing algebra, please use pencil &paper .

 

[Dr.Peterson]

= [k^2 (k + 1)^2 + 4 ( k + 1 )^3 ] / 4 -The LCD is 4
= [ ( k + 1 )^2 [ k^2 + 4 ( k + 1 ) ] ] / 4 - Factor

Now, the failure in my understanding is the part in step three where we go from the step explained by -The LCD is 4 and the step explained by -Factor.

My issue is that I have no idea where the ^3 is going in the factoring step. To me it just looks like the commutative property performed on the first term and then k^2 is somehow grouped with 4 ( k + 1 ) while dropping the ^3.

When you are told that something has been factored, you can best see that this is true by checking the result: Multiply the result and see that you get what they started with:

( k + 1 )^2 [ k^2 + 4 ( k + 1 ) ] = k^2 (k + 1)^2 + 4 ( k + 1 )^3​

This reveals what happened to the cube, and gives you a model for handling similar expressions in the future: Look for a common factor, and pull it out.

I feel ... maybe ... mislead a bit. I feel like this was poorly explained by the example.

I'd say you misled yourself, by letting your mind see irrelevant connections, rather than focus on what you were told (that the expression was factored). It takes experience to learn how to read mathematics (namely, as @khansaheb suggests, by doing the work yourself, in parallel to the author, on paper, perhaps writing out extra steps that the author didn't feel a need to write.