# Show the proof by induction in the given problem

• chwala
In summary, by adding the next term in the series, we can express the sum of cubes up to ##k+1## as the sum of cubes up to ##k## plus the cube of the next integer. This allows us to easily find a closed form expression for the sum of cubes.

#### chwala

Gold Member
Homework Statement
See attached
Relevant Equations
Induction
My interest is solely on the highlighted part in red...hmmmmmmm taken a bit of my time to figure that out...but i got it. Looking for any other way of looking at it;

I just realised that the next term would be given by;

##\dfrac{1}{4}(k+1)^2(k+2)^2-\dfrac{1}{4}k^2(k+1)^2##

##=\dfrac{1}{4}(k+1)^2\left[(k+2)^2-k^2\right]##

##=\dfrac{1}{4}(k+1)^2\left[k^2+4k+4-k^2\right]##

##=\dfrac{1}{4}(k+1)^2\left[4k+4\right]##

Therefore adding the ##k## to ##k+1## yields;

##\dfrac{1}{4}k^2(k+1)^2+ \dfrac{1}{4}(k+1)^2\left[4k+4\right]##

##=\dfrac{1}{4}(k+1)^2\left[k^2+4k+4\right]##

Then the rest of the step will follow...this was the only challenging part...any other way of looking at it would be nice. Cheers

Question
If i may ask, how did they get the ##(k+1)^3## on the first line?

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The line in red is just factoring ##\frac{1}{4}(k+1)^2##. The left term leaves a ##k^2##, the right term leaves a ##4(k+1)## which is less visually obvious because factoring fractions is tricky. You can avoid needing to notice clever manipulations. You can just totally expand the expression into a polynomial, and then take your target expression and expand it into a polynomial, and confirm the coefficients line up.

chwala said:
Question
If i may ask, how did they get the ##(k+1)^3## on the first line?

Where on the first line?

malawi_glenn said:
Where on the first line?
 On this line; ##\sum_{r=1}^{k+1} r^3= \dfrac{1}{4}(k+1)^2 +(k+1)^3##

chwala said:
 On this line; ##\sum_{r=1}^{k+1} r^3= \dfrac{1}{4}(k+1)^2 +(k+1)^3##

Hint, what can ## \displaystyle \sum_{r=1}^{k} r^3 + (k+1)^3## be written as?

what do you think "adding the next term" mean?

malawi_glenn said:
Hint, what can ## \displaystyle \sum_{r=1}^{k} r^3 + (k+1)^3## be written as?
View attachment 319251
what do you think "adding the next term" mean?

Seen it...you can check post ##1## and see that i know what 'the next term is'...therefore i will just go ahead and add it to previous term as follows;

##\dfrac{1}{4}k^2(k+1)^2+\dfrac{1}{4}(k+1)^2\left[4k+4\right]##

##\dfrac{1}{4}k^2(k+1)^2+4\left[\dfrac{1}{4}(k+1)^2(k+1)\right]##

##=\dfrac{1}{4}k^2(k+1)^2+(k+1)^2 (k+1)##

##\dfrac{1}{4}k^2(k+1)^2+(k+1)^3##

Cheers man!

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chwala said:
Cheers man!
It means this
## \displaystyle \sum_{r=1}^{k} r^3 + (k+1)^3 = 1 + 2^3 + 3^3 + \cdots + (k-1)^3 + k^3 + (k+1)^3 = \sum_{r=1}^{k+1} r^3 ##.
Thus, we can write
## \displaystyle \sum_{r=1}^{k+1} r^3 = \sum_{r=1}^{k } r^3 + (k+1)^3 = \dfrac{1}{4}k^2(k+1)^2 + (k+1)^3 ##
where we used the induction hypothesis in the last step.

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PeroK and chwala