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FP1 - proof by induction

I don't understand how to do proof by induction:
the question im stuck on is;

(the sum of) (3k+1) =1/2n(3n+5)

PLEASE HELP SOMEONE
Original post by Amyherdman
I don't understand how to do proof by induction:
the question im stuck on is;

(the sum of) (3k+1) =1/2n(3n+5)

PLEASE HELP SOMEONE


Have you tried it yet?
(edited 8 years ago)
Reply 2
Original post by edothero
Have you tried it yet?


yes but i dont know how to do it
Original post by Amyherdman
yes but i dont know how to do it


Remember the 4 steps.
Prove for n=1
Assume true for n=k
Let n=k+1 and prove
Conclusion

What step are you stuck on?
Reply 4
Original post by edothero
Remember the 4 steps.
Prove for n=1
Assume true for n=k
Let n=k+1 and prove
Conclusion

What step are you stuck on?


mainly assume true for n=k
Original post by Amyherdman
mainly assume true for n=k

As far as I'm aware, for this part you just need to write a statement saying you assume its true for n=k? There's not much more you need to do for this step..
Would be good if you could post your working out


@Zacken can take over here. I have a Physics exam to revise for :biggrin:
Reply 6
Original post by Amyherdman
mainly assume true for n=k


Yeah, all you need to do here is just write down Assume true for n=m, so that we have: i=1k(3i+1)=12k(3k+5)\sum_{i=1}^k (3i+1) = \frac{1}{2}k(3k+5)

That's it.

Original post by edothero


@Zacken can take over here. I have a Physics exam to revise for :biggrin:


Argh, trying to do S3 but I'll take over. :biggrin:
Reply 7
Original post by edothero
As far as I'm aware, for this part you just need to write a statement saying you assume its true for n=k? There's not much more you need to do for this step..
Would be good if you could post your working out

@Zacken can take over here. I have a Physics exam to revise for :biggrin:


thank you, ive done that but thought i had to do more to help me prove n=k+1
Original post by Amyherdman
mainly assume true for n=k


Write the statement in k rather than n, not too difficult.
Reply 9
Original post by Zacken
Yeah, all you need to do here is just write down Assume true for n=m, so that we have: i=1k(3i+1)=12k(3k+5)\sum_{i=1}^k (3i+1) = \frac{1}{2}k(3k+5)

That's it.



Argh, trying to do S3 but I'll take over. :biggrin:

what about n=k+1 then??
Reply 10
Original post by zetamcfc
Write the statement in k rather than n, not too difficult.


thats what i did but i thought there was more than that to do, how do you do n=k+1 then?
Reply 11
Original post by Amyherdman
what about n=k+1 then??


Then, for n=k+1, we want to prove that:

i=1k+1(3i+1)=??12(k+1)(3(k+1)+5)\displaystyle \sum_{i=1}^{k+1} (3i+1) =^{??} \frac{1}{2}(k+1)(3(k+1) + 5) - we want to prove this equality somehow, so let's start from the sum:

i=1k+1(3i+1)=i=1k(3i+1)+3(k+1)\displaystyle \sum_{i=1}^{k+1} (3i+1) = \sum_{i=1}^k (3i+1) + 3(k+1) - you've assumed something about the sum to k, maybe plug it in and do some factorising to prove what we want to prove?
Original post by Amyherdman
thats what i did but i thought there was more than that to do, how do you do n=k+1 then?


@Zacken will help, got stuff to do.
Hint: The sum to k+1 = The sum to k + the (k+1)th term. Now substitute for the sum to k with the inductive hypothesis (the assumption that the statement is true for n=k).

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