# Thread: Inductive reasoning

1. ## Inductive reasoning

Sort of lost on how to continue...
Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

n = 1 is true
assume k1 is true? (4k - 1) = k(2k + 1)
now to prove k + 1

[4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

Would someone mind showing me where to go from here?

2. ## Ok

hi-

i think u are on the right track.
continue.

Just a step and u are there!

3. Originally Posted by Såxon
Sort of lost on how to continue...
Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

n = 1 is true
assume k1 is true? (4k - 1) = k(2k + 1)
now to prove k + 1

[4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

Would someone mind showing me where to go from here?
you mean

$\sum_{i=1}^{n} (4i-1) = n(2n+1)$

let $P(k):\sum_{i=1}^{k} (4i-1) = k(2k+1)$

$P(1)$ is true and $P(2)$ is true suppose that P(k) is true lats try

$P(k+1):\sum_{i=1}^{k+1} (4i-1) ={\color{blue} (k+1)(2(k+1)+1)}$ true or not

$\sum_{i=1}^{k+1}(4i-1)={\color{red}\sum_{i=1}^{k}(4i-1)}+(4(k+1)-1)$

and the red term equal $k(2k+1)$ from P(k) so you will have

$\sum_{i=1}^{k+1}(4i-1)=k(2k+1)+(4(k+1)-1)=2k^2+k+(4k+3)=2k^2+5k+3=(k+1)(2k+3)$ and this equal the blue term the proof ends

4. By the way, "proof by induction", which is what is meant here, and "inductive reasoning" are not at all the same thing!

5. Originally Posted by Såxon
Sort of lost on how to continue...
Prove that the following formula is true for all n using inductive reasoning 3 + 7 + 11 + .... +(4n - 1) = n(2n + 1)

n = 1 is true
assume k is true: 3 + 7 + 11 + .... +(4k - 1) = k(2k + 1)
now to prove k + 1

3 + 7 + 11 + .... +(4k - 1)+[4(k+1) - 1] + (4k - 1) = k(2k + 1) + [4(k+1) - 1]

Would someone mind showing me where to go from here?
I've added some parts that should be there -- I can't tell if you omitted them for less typing, or weren't sure that they should be there.

You want to show that the last thing you typed: k(2k + 1) + [4(k+1) - 1] is equal to (k+1)(2(k+1) + 1), for the inductive statement to be true.

Show that as Amer did:
$k(2k+1)+(4(k+1)-1)=2k^2+k+(4k+3)=2k^2+5k+3=(k+1)(2k+3)$