Is anyone good at mathematical induction? I'm so confused!

- Meehan98

Is anyone good at mathematical induction? I'm so confused!

- schrodinger

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- Meehan98

##### 1 Attachment

- Meehan98

I know that the first step is to make n=1 in the equation, but I don't understand the "n=k+1" part. Do I just input "k+1" every time there's an "n" in the equation?

- welshfella

Proof by induction first attempts to prove that the formula is true for k = 1. The process assumes that the result is true for n = k. Then the (k+1 )th term is added. If the resulting expression is the same as the expression for the sum for k terms except that k is replaced by (K+1) then this constitutes the proof.

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- welshfella

* true for n = 1

- welshfella

- because if iot is true for n = 1 then it must be true for n = 2 , 3 etc

- Meehan98

So, the first step would give you 1+3+5+...+(1)=1

- Meehan98

And then if n=k, you get:
1+3+5+...+(2k-1)=k^2

- welshfella

No the first step is what is the sum of n terms if n = 1?
1 = first term so sum of 1 term = 1
n^2 = 1^2 = 1 = sum of 1 term
so true for n = 1

- welshfella

yes that is the sum of k terms

- welshfella

now add the (k + 1)th term

- welshfella

what is the formula for (k+1)th term
kth term = 2k - 1 right?

- Meehan98

Ok, is this looking right so far?
1+3+5+...(2k-1)+(k+1)=k^2+(k+1)

- Meehan98

Because I add the (k+1)th term to each side?

- welshfella

no the (k+1)th term is
(2(k + 1) - 1)
|dw:1443022249395:dw|

- welshfella

the k is replaced by (k + 1)

- Meehan98

Oh, okay that makes sense. Do we do the same thing to the other side of the equation?

- Meehan98

If so, the equation I have so far is 1+3+5+...+(2k+1)-1)=(k+1)^2

- welshfella

yes the right side of the equation now becomes k^2 + 2(k + 1) - 1
which if you expand comes to (k +1)^2

- welshfella

- and thats your proof
(k + 1)^2 is k^2 with the k replaced by (k +1)

- Meehan98

Okay, would we leave (2k-1) in the equation, because the first option looks like the correct answer but it still has (2k-1) in it?

- welshfella

yes leave 2k-1 there the first option is the correct one

- welshfella

2k-1 is the kth term
you need to find the general formula for any term in order to do the proof

- welshfella

its a bit tricky to learn Induction . Practice will be the best way to grasp it.

- Meehan98

Thank you! Yes, it's a lot to take in at first. Do you have time to help me with one more?

- welshfella

i think so

- Meehan98

##### 1 Attachment

- welshfella

First plug n = 1 into the right hand formula and see if it gives 1

- Meehan98

So, my first step is to make n=1 and I get 1^3=1(4)/4 and this is true.

- welshfella

yea good work
now let n = k
and then write formula with k

- Meehan98

My next step is to make n=k and I get: 1^3+2^3+3^3+...+k^3=k^2(k+1)^2/4

- welshfella

yes exactly

- welshfella

now add (the (k + 1)th term to both sides

- Meehan98

Now i replace k with (k+1)
(k+1)^3=(k+1)(k+1)+1)^2/4

- welshfella

No you ADD (k + 1)^3 to both sides
You are aiming for the formula that you have written but havent got it yet

- welshfella

RHS is k^2 + (k + 1)^2 + (k + a)^3
-------------
4

- welshfella

you now have to simplify this

- welshfella

sorry its not a plus there
correct is
k^2 (k + 1)^2
------------ + (k + 1)^3
4

- Meehan98

This is so confusing; I thought that you would multiply by 4 to each side to get rid of the denominator, but you don't. I don't know the first step in order to simplify this. Sorry.

- welshfella

make this into one fraction

- welshfella

no you multiply the (k + 1)^3 by 5
so its
k^2(k + 1)^2 + 4(k + 1)^3
----------------------
4

- welshfella

now take out the common factor (k + 1)^2
= (k + 1)^2( k^2 + 4k + 4)
------------------
4

- welshfella

* i should have said multiply by 4

- welshfella

do you follow the above?

- Meehan98

Yes, so you get: (k+1)^2 (k+2)^2/4

- Meehan98

Then you get: (k+1)^2 ((k+1)+1)^2/4 which is the original n^2(n+1)^2/4

- welshfella

exactly
which is the formula for sum of k terms with the k replaced by k+1
So if the formula is trye for n = k then its also true for n = k + 1.

- welshfella

we've also shown that its true for n = 1 so it must be true for n = 2,3 etc True for all whole number values of n.

- Meehan98

Thank you! You've been wonderful help!

- welshfella

for practice try a relatively easy one
show by induction
1+2+3+ + n = (n/2)(n + 1)

- welshfella

yw

- Meehan98

Okay, I'm close but not there yet. After making the n=1 and n=k I got: k=k/2(k+1)
now I add (k+1)th term and I got:
(k+1)=k/2 (k+1)+(k+1), but that would equal (k+1)=k/2 2(k+1) so I'm doing something wrong.

- welshfella

we have (k/2)(k + 1) + k + 1
take out the common factor (k + 1)
= (k + 1( (k/2) + 1)
= ( k + 1) ( k + 2)
----
2
= (k + 1) (( k + 1) + 1)
-----
2
which is the formula for sum of k terms with k replaced by (k + 1)

- welshfella

you have to 'juggle ' the terms about a bit
- this is the trickiest part

- welshfella

I thought this was straight forward but it wasnt!

- welshfella

- a good idea is to write down what you are aiming for before you attempt the algebra

- Meehan98

Okay, thank you for all of your help!

- welshfella

yw

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