Question

Prove using mathematical induction that 1+4+7+10+...+(3n-2)=(n/2)(3n-1)

Answer 1

Let's assume $$1+4+7+10+\dots+(3n-2)=n(3n-1)/2$$for some positive integer \(n\). Now we need to show that this is enough to demonstrate our statement *also* true for \(n+1\).

Answer 2

Since we want to end up deriving$$1+4+7+10+\dots+(3n-2)+(3(n+1)-2)=(n+1)(3(n+1)-1)/2$$we can see want to add a \(3(n+1)-2\) term to either side:$$1+4+7+10+\dots+(3n-2)+(3(n+1)-2)=n(3n-1)/2+(3(n+1)-2)$$

Answer 3

can you figure out how to go from \(n(3n-1)/2+(3(n+1)-2)\) to \((n+1)(3(n+1)-1)/2\)

Answer 4

I'm a little confused on the process of how you go from the written problem to the last statement you made

Answer 5

the key is that we want to show that it holds for some positive integer \(n\) it'll also hold for the next positive integer \(n+1\); then, if we can show it holds for \(n=1\), it follows it also holds for \(n=2\), \(n=3\), and so forth for all positive integers \(n\).

Answer 6

Okay that makes a little more sense, i guess my next question would be how do you know to add 3(n+1)-2 to both sides

Answer 7

@oldrin.bataku

Answer 8

well the idea is to consider the claim for \(n+1\) would mean:$$1+4+7+10+\dots+(3n-2)+(3(n+1)-2)=(n+1)(3(n+1)-1)/2$$so all we did was replace everywhere we saw \(n\) with \(n+1\). to prove the statement holds for \(n+1\) is to prove the above equation starting from the case for \(n\) i.e.$$1+4+7+10+\dots+(3n-2)=n(3n-1)/2$$so clearly we see our left-hand side needs a \(3(n+1)-2\) which is why we added that

Answer 9

Ohhhhhhhhhhhhhh that makes more sense now! So once you add it what do you do then?

Answer 10

@oldrin.bataku when you complete the problem what should your final answer look like, i think that's where I'm confused, you don't have to give me an answer, just like a how to:) thank you soooooo much for your help

Answer 11

sorry, I've been eating! well, once you add it, the key is just to play with the right-hand side until it looks like \((n+1)/(3(n+1)-1)/2\)

Answer 12

so currently our right-hand side is:$$n(3n-1)/2+3(n+1)-2$$

Answer 13

notice:$$n(3n-1)/2+3(n+1)-2=\frac{n(3n-1)+2(3(n+1)-2)}2=\frac{3n^2-n+6n+6-4}2$$

Answer 14

(we just found a common denominator and combined numerators)

Answer 15

now combine like terms in our numerator:$$3n^2+5n+2$$and factor:$$3n^2+5n+2=3n^2+3n+2n+2=3n(n+1)+2(n+1)=(n+1)(3n+2)$$

Answer 16

lastly we rewrite \(3n+2=3n+3-1=3(n+1)-1\) so our RHS is finally:$$(n+1)(3(n+1)-1)/2$$ and we've shown that if our statement holds for \(n\) it also necessarily holds for the next number \(n+1)\). all we have to do now is show it holds for \(n=1\):$$1=1(3(1)-1)/2=2/2=1$$and so it then must hold for \(n=2\); from there it holds for \(n=3\), and then \(n=4\), etc. -- for all positive integers

Answer 17

now combine like terms in our numerator:$$3n^2+5n+2$$and factor:$$3n^2+5n+2=3n^2+3n+2n+2=3n(n+1)+2(n+1)=(n+1)(3n+2)$$

Answer 18

so we are able to finish with (3n+2)(n+1)

Answer 19

Should I factor that out, or leave it in that form?

Answer 20

but doesnt the original problem say (n/2)(3n-1)

Answer 21

@oldrin.bataku thanks for all your help

Answer 22

Okay I have figured out the above questions, but I'm actually confused with what happened to the left side

Answer 23

that was just our numerator factored; keep the denominator so \((n+1)(3(n+1)-1)/2\)