Question

prove the following by mathmatical induction:
3+7+11+...+(4-1)=n(2n+1)

Answer 1

4=n(2n+1) ?

Answer 2

oops, the left side is (4n-1)

Answer 3

\[\large 4n-1=n(2n+1)\] solving for n?

Answer 4

Oh wait nvm.... not solving for n.

Answer 5

Yeah.... how did you solve that now...lol

Answer 6

any ideas lol

Answer 7

oh i see, because n = n+1 you could easily substitute the "in for 4(n=1) which is basically an and an - 1 = 4(n+1)-1

so it's saying an + an-1 = 2an...... i think.

Answer 8

im giving up i need to look up what induction is!

Answer 9

ahh...

Answer 10

oh HEY! i think i got it!

Answer 11

yeah?

Answer 12

LHS: RHS
let S be the sequence
for n = 1
4(1)-1=3 : 1*2*1+3 =3
s(n+1)=(n+1)(2*(n+1)+1
=2n^2+5n+3
=2n^2+n+4n+3
=n(2n+1)+(4n+3)
=S(n)+4(n+1)-1.... done

Answer 13

does that make sense?

Answer 14

Yeah... thats how i went about solving it but i didn't want it to be confusing with how you were explaining it.

i got the 2n^2+5n+3

Answer 15

oh i dunno how to do induction lol i just looked up its meaning

Answer 16

Its kind of recursive in the sense that there's always that (4n+1) there somewhere in the problem.

Answer 17

show me the right formal for induction i need to see

Answer 18

format*

Answer 19

Yes the idea is if its true for the base case and you can show that if its true for n then its true for n + 1. You can reach any case by applying the implication starting from the base case over and over until you reach the given index.

Answer 20

Whoah, too complicated. :|

Answer 21

show its true for 1
assume its true for some n
show its true(with that assumption) for n+1

Answer 22

I can write it out and scan it but it might take me a min, have you solved it yet?
I need to use sigma notation and that would take me to long on here.

Answer 23

scan it? lmao xD

Answer 24

im pretty sure my solution works, but i dont think its proper format

Answer 25

This is how induction works. Why does induction give us that a given property is true for all \(n\). Lets say \(P(n)\) is a predicate which is true if a given property holds for \(n\). So if we can show that \(P(n)\) is true for \(P(1)\) and that \(P(n) \implies P(n + 1)\) then we can reach any given \(n\) by recursively applying the implication starting from the base case \(1\). So why does it hold for 3 just as an example? Because \(P(1) \implies P(2)\) and therefore \(P(2) \implies P(3)\).

Answer 26

Yes and for the original poster, if you carefully read my discussion on induction you will understand why these proofs make sense to begin with.

Answer 27

its dominoes, if one falls and then the next falls, then the rest will fall.

Answer 28

Actually that's a nice visual description of it.

Answer 29

@comput313 let me know if you have any questions about the proof.

Answer 30

ok it is true for n =1
assume sigma from 1 to n (4i-1) = n(2n+1)
then sigma from 1 to n+1 (4i-1) = 4(n+1) + sigma from 1 to n of (4i-1)
= 4(n+1) -1+n(2n+1) = 2n^2+5n+3 = (n+1)(2(n+1) + 1)
thus by induction it is true for all n

Answer 31

sorry typo fixed

Answer 32

You can use latex here =P. $$\displaystyle\large\sum_{i=1}^n$$

Answer 33

I know the math not the latek:)

Answer 34

\[\sum_{i = 1}^{n} ( 4i-1) and \sum_{i=1}^{n+1} (4i-1)\]

Answer 35

@zzr0ck3r rewriting.\[T \to n=1\]\[\large \sum_{n=1}^{n}(4n-1) = n(2n+1)\]\

Answer 36

Haha aww.... ok. I shall let you continue. LaTex is fun.

Answer 37

Indeed, I used LaTeX for all of my problem set solutions for a course this year.

Answer 38

\[\sum_{i =1}^{n} (4i-1) = n(2n+1)\]
then
\[\sum_{i=1}^{n+1}(4i-1) = 4(n+1)-1+\sum_{i=1}^{n}(4i-1) = \]
4(n+1) -1+n(2n+1) = 2n^2+5n+3 = (n+1)(2(n+1) + 1)
qed.

Answer 39

@comput313

Answer 40

No, he's probably long gone.

Answer 41

bah

Answer 42

oh thanks zzrocker :)

Answer 43

now i understand induction

Answer 44

\[\large \sum_{n=1}^{n}(4n-1) = n(2n+1)\]\[\large \sum_{n=1}^{n+1}(4n-1)=4(n+1)+\sum_{n=1}^{n}(4n-1)\]\[\large = 4(n+1) -1+n(2n+1) = 2n^2+5n+3 = (n+1)(2(n+1) + 1)\]

Answer 45

mwuahahaha

Answer 46

good deal. you will need it for any 300+ level math class.

Answer 47

Super Saiyan :P Jk..

Answer 48

well iam only 12 years old, so i have time

Answer 49

12 in elephant years

Answer 50

let n = 1
then for sure
3 = 3
so assume its true for some n
thus
\[\sum_{i=1}^{n}(4i-1) = n(2n+1)\]
consider n+1
then
\[\sum_{i=1}^{n+1}(4i-1) = 4(n+1)-1+\sum_{i=1}^{n}(4i-1)\]
we know from out assumption about n =1 that this is equal to
\[4(n+1)-1+n(2n+1) = (n+1)(2(n+1)+1)\]
thus by induction we have shown that if it is true for n it is true for n+1
and since its true for n =1
it follows that it is true for all n in N.

Answer 51

yes lots of time.:)

Answer 52

omg i forgot the "-1" after "4(n-1)" -_____- oh mygoodness.