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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?
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.
* true for n = 1
- because if iot is true for n = 1 then it must be true for n = 2 , 3 etc
So, the first step would give you 1+3+5+...+(1)=1
And then if n=k, you get: 1+3+5+...+(2k-1)=k^2
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
yes that is the sum of k terms
now add the (k + 1)th term
what is the formula for (k+1)th term kth term = 2k - 1 right?
Ok, is this looking right so far? 1+3+5+...(2k-1)+(k+1)=k^2+(k+1)
Because I add the (k+1)th term to each side?
no the (k+1)th term is (2(k + 1) - 1) |dw:1443022249395:dw|
the k is replaced by (k + 1)
Oh, okay that makes sense. Do we do the same thing to the other side of the equation?
If so, the equation I have so far is 1+3+5+...+(2k+1)-1)=(k+1)^2
yes the right side of the equation now becomes k^2 + 2(k + 1) - 1 which if you expand comes to (k +1)^2
- and thats your proof (k + 1)^2 is k^2 with the k replaced by (k +1)
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?
yes leave 2k-1 there the first option is the correct one
2k-1 is the kth term you need to find the general formula for any term in order to do the proof
its a bit tricky to learn Induction . Practice will be the best way to grasp it.
Thank you! Yes, it's a lot to take in at first. Do you have time to help me with one more?
i think so
First plug n = 1 into the right hand formula and see if it gives 1
So, my first step is to make n=1 and I get 1^3=1(4)/4 and this is true.
yea good work now let n = k and then write formula with k
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
now add (the (k + 1)th term to both sides
Now i replace k with (k+1) (k+1)^3=(k+1)(k+1)+1)^2/4
No you ADD (k + 1)^3 to both sides You are aiming for the formula that you have written but havent got it yet
RHS is k^2 + (k + 1)^2 + (k + a)^3 ------------- 4
you now have to simplify this
sorry its not a plus there correct is k^2 (k + 1)^2 ------------ + (k + 1)^3 4
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.
make this into one fraction
no you multiply the (k + 1)^3 by 5 so its k^2(k + 1)^2 + 4(k + 1)^3 ---------------------- 4
now take out the common factor (k + 1)^2 = (k + 1)^2( k^2 + 4k + 4) ------------------ 4
* i should have said multiply by 4
do you follow the above?
Yes, so you get: (k+1)^2 (k+2)^2/4
Then you get: (k+1)^2 ((k+1)+1)^2/4 which is the original n^2(n+1)^2/4
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.
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.
Thank you! You've been wonderful help!
for practice try a relatively easy one show by induction 1+2+3+ + n = (n/2)(n + 1)
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.
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)
you have to 'juggle ' the terms about a bit - this is the trickiest part
I thought this was straight forward but it wasnt!
- a good idea is to write down what you are aiming for before you attempt the algebra
Okay, thank you for all of your help!