Induction (is the devil) help please ;A; (1 Viewer)

[tooyoungforthis]

hi everyone,

I was wondering if I could get some help for induction, it's a fairly simple question but I'm having a hard time with it.

1. Prove by M.I that the sum of consecutive odd positive integers is divisible by 4.
(I can't get the equation you start with )

and as for inequalities...
I don't understand the steps and how it turns out to be an equal sign or how you multiply both sides with the number and it's just a mess >.< can someone please break it down for me?

you can use : 3ⁿ >/ 2ⁿ + n for all n>/1 as an example

thanks in advance!

 

[SpiralFlex]

First question specify how many consecutive integers you are after

Second question

 

[tooyoungforthis]

First question specify how many consecutive integers you are after

Second question

the first question didn't have any specification, I got it from my text book and it says just that.

and thank you for the second question! I'm starting to understand it now, I've just got to practice a lot more
(btw, I love your handwriting)

 

[phoenix159]

hi everyone,

I was wondering if I could get some help for induction, it's a fairly simple question but I'm having a hard time with it.

1. Prove by M.I that the sum of consecutive odd positive integers is divisible by 4.
(I can't get the equation you start with )

and as for inequalities...
I don't understand the steps and how it turns out to be an equal sign or how you multiply both sides with the number and it's just a mess >.< can someone please break it down for me?

you can use : 3ⁿ >/ 2ⁿ + n for all n>/1 as an example

thanks in advance!

There must be something missing in question 1 otherwise it wont work

 

[phoenix159]

You can start by finding the sum of consecutive odd integers:

1 + 3 + 5 + ... + (2n - 1)

S_n = n/2 (2a + (n-1)d) = n/2 (2x1 + (n - 1)x2) = n/2 (2 + 2n - 2) = n * n = n²

But this is not always divisible by 4

 

[obliviousninja]

First question specify how many consecutive integers you are after

Second question

typical neck destroyer

 

[SpiralFlex]

 

[SpiralFlex]

As an exercise try proving the result without mathematical induction

 

[SpiralFlex]

The result is for the sum of two consecutive odd integers

 

[SpiralFlex]

More generally you may want to see if you can prove the sum of n consecutive odd integers is divisible by n.

 

[phoenix159]

The result is for the sum of two consecutive odd integers

Then it would just be (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is obviously divisible by 4

 

[phoenix159]

More generally you may want to see if you can prove the sum of n consecutive odd integers is divisible by n.

1 + 3 + 5 + ... + (2n + 1) = n/2 (2a + (n-1)d) = n/2(2 + (n-1)2) = n(1 + n - 1) = n^2, which is divisible by n

 

[SpiralFlex]

Then it would just be (2n + 1) + (2n + 3) = 4n + 4 = 4(n + 1), which is obviously divisible by 4

Yup

 

[SpiralFlex]

But unfortunately, HSC papers like to get you to prove via induction (which can be tedious). For example a common induction problem is prove that is divisible by 6.

But it can be achieved using one a couple of lines (or less). Equivalent of proving for some integer

 

[Carrotsticks]

But unfortunately, HSC papers like to get you to prove via induction (which can be tedious). For example a common induction problem is prove that is divisible by 6.

But it can be achieved using one a couple of lines (or less). Equivalent of proving for some integer

 

[SpiralFlex]

I think this question was also in bos trials^

 

[RealiseNothing]

Alternatively is just the sum of the first 'n' triangular numbers.