expanding some polynomials: [k^2+k(2k+1)]/6 + [(k+1)^2]/1

[Havie]

Hello all,
It seems my algebra skills have escaped me because im having some troubles.
What I need to do is make the green statement look like the red statement in this photo (prove somethings true)




What I cant figure out is how to expand the red eqn at all..

particularly
(k+1)*((k+1)+1)

if you could help me here it would be greatly appreciated, thanks!

 

[Harry_the_cat]

Hello all,
It seems my algebra skills have escaped me because im having some troubles.
What I need to do is make the green statement look like the red statement in this photo (prove somethings true)

View attachment 10507
https://i.imgur.com/HKdr3Jh.jpg

What I cant figure out is how to expand the red eqn at all..

particularly
(k+1)*((k+1)+1)

if you could help me here it would be greatly appreciated, thanks!

(k+1)*((k+1)+1) = (k+1))(k+2) = ...

 

[JeffM]

Your first problem if you are supposed to prove

\(\displaystyle n \in \mathbb Z \text { and } n \ge 1 \implies \displaystyle \sum_{j=11}^n j^2 = \dfrac{n(n + 1)(2n + 1)}{6}\),

is that you do not appear to have solved the case if n = 1. It is admittedly not hard to show that

\(\displaystyle n = 1 \implies \displaystyle \sum_{j=1}^n j^2 = 1^2 = 1 = \dfrac{6}{6} \implies\)

\(\displaystyle n = 1 \implies \displaystyle \sum_{j=1}^n j^2 = \dfrac{2 * 3}{6} = \dfrac{1 * 2 * 3}{6} = \dfrac{n(n + 1)(2n + 1)}{6}.\)

Having done that absolutely necessary first step, you can obviously say

\(\displaystyle \exists \text { set } \mathbb K \text { such that } k \in \mathbb K \implies\)

\(\displaystyle k \in \mathbb Z, \ k \ge 1, \text { and } \displaystyle \sum_{j=1}^k j^2 = \dfrac{k(k + 1)(2k + 1)}{6}.\)

Now before you go running off fussing with polynomials, you need to think about what you must show for k+1.

\(\displaystyle \displaystyle \sum_{j=1}^{k+1} j^2 = WHAT?.\)

What you are always trying to do in a proof by induction is to use what you know about k.

\(\displaystyle \displaystyle \sum_{j=1}^{k+1} j^2 = \left ( \sum_{j=1}^k j^2 \right ) + (k + 1)^2 =\)

\(\displaystyle \dfrac{(k^2 + k)(2k + 1)}{6} + k^2 + 2k + 1 = \dfrac{2k^3 + 3k^2 + k}{6} + \dfrac{6k^2 + 12k + 6}{6} =\)

\(\displaystyle \dfrac{2k^3 + 9k^2 + 13k + 6}{6}.\)

Now what must you do with that mess? You must show that

\(\displaystyle \dfrac{2k^3 + 9k^2 + 13k + 6}{6} = \dfrac{(k + 1)\{(k + 1) + 1\}\{2(k + 1) + 1\}}{6}.\)

Now you can make life easier for yourself by seeing that

\(\displaystyle \dfrac{(k + 1)\{(k + 1) + 1\}\{2(k + 1)\} + 1\}}{6} = \dfrac{(k + 1)(k + 2)(2k + 3)}{6}.\)

So start by showing that

\(\displaystyle 2k^3 + 9k^2 + 13k + 6 = (k + 1)(k + 2)(2k + 3).\)

How might you do that?

 

[ksdhart2]

The expression you've highlighted in red is a bit nasty, but truly it's no different from any other expression you've expanded in the past. My best guess is that you're getting intimidated and can't figure it out because you're taking in the whole thing at once and it's just too much. To counteract this, let's break it down and analyze it part-by-part, performing just one step at a time. First, write down the unsimplified version, just so we know what we're dealing with. I've separated out the nested grouping symbols instead of using only parentheses, and inserted the implied multiplication symbols:

\(\displaystyle \dfrac{(k+1) \cdot [(k+1)+1] \cdot \{[2(k+1)]+1\}}{6}\)

Now, the first term seems okay on its own. I see no immediate simplification/expansion to be done, so let's isolate the middle term and see what we can do with it:

\(\displaystyle [(k+1)+1]\)

Looking at it, it should be obvious why the redundant parentheses can be removed simplifies down to k + 2. Okay, so what about the final term? What can we do there?

\(\displaystyle \{[2(k+1)]+1\}\)

Do you see that we can expand the 2(k+1) part into 2k + 2, thus making the whole term into 2k + 3? It really seems like we're getting somewhere! Since we've simplified/expanded each individual term, let's put everything back in and see where we stand now:

\(\displaystyle \dfrac{(k+1) \cdot (k+2) \cdot (2k+3)}{6}\)

Try continuing from here by multiplying all the terms together. What do you get? How does that relate to the expression highlighted in green?

 

[Havie]

The expression you've highlighted in red is a bit nasty, but truly it's no different from any other expression you've expanded in the past. My best guess is that you're getting intimidated and can't figure it out because you're taking in the whole thing at once and it's just too much. To counteract this, let's break it down and analyze it part-by-part, performing just one step at a time. First, write down the unsimplified version, just so we know what we're dealing with. I've separated out the nested grouping symbols instead of using only parentheses, and inserted the implied multiplication symbols:

\(\displaystyle \dfrac{(k+1) \cdot [(k+1)+1] \cdot \{[2(k+1)]+1\}}{6}\)

Now, the first term seems okay on its own. I see no immediate simplification/expansion to be done, so let's isolate the middle term and see what we can do with it:

\(\displaystyle [(k+1)+1]\)

Looking at it, it should be obvious why the redundant parentheses can be removed simplifies down to k + 2. Okay, so what about the final term? What can we do there?

\(\displaystyle \{[2(k+1)]+1\}\)

Do you see that we can expand the 2(k+1) part into 2k + 2, thus making the whole term into 2k + 3? It really seems like we're getting somewhere! Since we've simplified/expanded each individual term, let's put everything back in and see where we stand now:

\(\displaystyle \dfrac{(k+1) \cdot (k+2) \cdot (2k+3)}{6}\)

Try continuing from here by multiplying all the terms together. What do you get? How does that relate to the expression highlighted in green?

thanks,
this is exactly what I needed. After 2 days of 6 hours of discrete math i had forgotten routine algebra. Other post above this was just way over the top, im sure it can be turned into sequences but ive never learned those, just briefly know of them.