Question

Use mathematical induction to prove the statement is true for all positive integers n, or show why it is false.

1^2 + 4^2 + 7^2 + ... + (3n-2)^2 = n(6n^2-3n-1)/2

this is so hard, I need help :(

Answer 1

@amistre64 do you know this by any chance??

Answer 2

What part of the inductive process are you stuck at? Base Step? Inductive Step?

Answer 3

ok I know that the first step is to have n=1 and when I did that it came out as true. But now I'm stuck on the S(k+1) step and on. It's just really confusing to me

Answer 4

Alright, its good that you know the process. Your base step is done. Now you need to assume that:\[1^2+2^2+\cdots +k^2=\frac{1}{2}k(6k^2-3k-1)\]for some fixed natural number k. This is called your Inductive Hypothesis. What we need to prove (making sure to use the Inductive Hypothesis) is that:\[1^2+2^2+\cdots +k^2+(k+1)^2 =\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 5

Does what I'm doing make sense so far? (like how I got those equations, etc.)

Answer 6

oh woah woah, my bad, I didn;t fully read the problem, let me correct some stuff.

Answer 7

Inductive Hypothesis is:\[1^2+4^2+7^2+\cdots +(3k-2)^2=\frac{1}{2}k(6k^2-3k-1)\]

Answer 8

What we want to prove is:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]\[=\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 9

Sry, I saw the squares and automatically assumed the problem was the sum of the first n squares.

Answer 10

wait can u break this down for me I am still really confused! where did the k^2 and the 1/2k come from in the beginning?

Answer 11

Ignore my earlier post, just look at the last three. Are there any questions where those equations came from?

Answer 12

ok it's making more sense but just one question so far in 12+42+72+⋯+(3k−2)2+(3(k+1)−2)2
=12(k+1)(6(k+1)2−3(k+1)−1) where did u get the second +3(k+1)-2)^2

Answer 13

That is the S(k+1) that you brought up earlier. The statement we are trying to prove is:\[1^2+4^2+7^2+\cdots +(3n-2)^2 =\frac{1}{2}n(6n^2-3n-1)\] This is S(n). S(k) is the inductive hypothesis, so I replace n with k. S(k+1) is what I want to prove, so I replace n with k+1.

Answer 14

ahh okay I'm beginning to understand that part now, so what I'm wondering is since in the original we have (3n-2)^2 so if we were just replacing n with k wouldn't we just have (3k+1-2)^2? Because the way you did it you left it as but added (3k-2)^2+(3k+1-2)^2

Answer 15

are you just suppose to keep it as (3k-2)^2 and then add it replacing it the second time?

Answer 16

Ah, thats to help us actually solve the problem. To see why i left (3k-2)^2 in the S(k+1), lets look at some specific values of n, and see what the left side of the equation becomes.

If n =1, then 3(1)-2=1, so the left side is:\[1^2\]
If n=2, then 3(2)-2=4, so the left side is:\[1^2+4^2\]
If n =3, then 3(3)-2=7, so the left side is:\[1^2+4^2+7^2\]

Notice that each time you increase n by 1, all you're really doing is adding a new term to the end of the summation. That is why when I wrote down S(k+1), I left the (3k-2)^2 in the summation. Its still there, and (3(k+1)-2)^2 gets tacked on to the end.

Answer 17

Its going to become super apparent why I left (3k-2)^2 in the summation when we start the inductive process.

Answer 18

ahhh okay that makes way more sense now!! thank you for clearing it up :) what's the next step after setting that up? the inductive process?

Answer 19

Right, now we actually have to prove that:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]
\[=\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 20

But we have to make sure to use the Inductive Hyptothesis. That should be the first thing what we do. We are assuming that:\[1^2+4^2+7^2+\cdots +(3k-2)^2=\frac{1}{2}k(6k^2-3k-1)\]is true. Somehow we need to mathematically manipulate:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]and obtain:\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]

Answer 21

This is the reason I left (3k-2)^2 in the summation. Note that:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]\[=\left(1^2+4^2+7^2+\cdots +(3k-2)^2\right)+(3(k+1)-2)^2\](all i did was group together all the terms but the last)

Answer 22

By the inductive hypothesis, we know that:\[1^2+4^2+7^2+\cdots +(3k-2)^2=\frac{1}{2}k(6k^2-3k-1)\]So we can substitute this in for whats in the bigger parenthesis. Does this make sense?

Answer 23

So performing this substitution gives us:\[\left(1^2+4^2+7^2+\cdots +(3k-2)^2\right)+(3(k+1)-2)^2=\]\[\left(\frac{1}{2}k(6k^2-3k-1)\right)+(3(k+1)-2)^2\]This step was really important, so if anything is fuzzy say so.

Answer 24

wait okay. so I see that you substituted the 1/2k(6k^2−3k−1) in where the 1^2+4^2+7^2+⋯+(3k−2)^2 used to be, but why did you do that?? I'm sorry if that's a dumb question :( like I see what you're doing, I'm just a little bit unclear

Answer 25

No no, questions are good, especially at this part. By substituting, I am using the inductive hypothesis. If you never use the inductive hypothesis, then you never did an inductive proof.

Recall, we are trying to turn:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]into:\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]Now, without using the inductive hypothesis, that might be a pretty hard task (not impossible, but lengthy for sure). By using the inductive hypothesis, look at how much closer we are to our goal! We just need to formally manipulate:\[\frac{1}{2}k(6k^2-3k-1)+(3(k+1)-2)^2\]to reach our goal.

Answer 26

Any more questions for anything up till now? The rest of the problem is algebra.

Answer 27

Alright, to finish this problem, I'm going to fully expand both:\[\frac{1}{2}k(6k^2-3k-1)+(3(k+1)-2)^2\] and :\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]and show that they are equal. This will complete the proof by induction.

Answer 28

Note that:\[\frac{1}{2}k(6k^2-3k-1)+(3(k+1)-2)^2\]\[=\frac{1}{2}(6k^3-3k^2-k)+(3k+1)^2\]\[=3k^3-\frac{3}{2}k^2-\frac{1}{2}k+9k^2+6k+1\]\[=3k^3+\frac{15}{2}k^2+\frac{11}{2}k+1\]All i did here was algebraic manipulation. Distributing, squaring, collecting like terms, etc.

Answer 29

(If any of my steps are wrong let me know, i tend to make simple mistakes lol)

Answer 30

Now Im going to fully expand:\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\] and I hope to get the same thing as in the last post.

Answer 31

Here we go (fingers crosses):\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]\[=\frac{1}{2}(6(k+1)^3-3(k+1)^2-(k+1))\]\[=\frac{1}{2}(6(k^3+3k^2+3k+1)-3(k^2+2k+1)-k-1)\]\[=\frac{1}{2}(6k^3+18k^2+18k+6-3k^2-6k-3-k-1)\]\[=\frac{1}{2}(6k^3+15k^2+11k+2)\]\[=3k^3+\frac{15}{2}k^2+\frac{11}{2}k+1\]

Answer 32

So we did get the same thing, and we have (finally) proved that:\[1^2+4^2+7^2+\cdots +(3k-2)^2+(3(k+1)-2)^2\]is the same as:\[\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)\]This completes the inductive step of a proof by induction.

Answer 33

Since we have completed both the base step and the inductive step, we have fully proven that:\[1^2+4^2+7^2+\cdots +(3n-2)^2=\frac{1}{2}n(6n^2-3n-1)\]for all natural numbers n.

Answer 34

@joemath314159 Oh my gosh thank you so much!!! it really makes more sense now!