- #1

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- 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

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

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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