# Inductive reasoning

• Jun 8th 2009, 02:41 AM
Såxon
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?
• Jun 8th 2009, 04:52 AM
Chandru1
Ok
hi-

i think u are on the right track.
continue.

Just a step and u are there!
• Jun 8th 2009, 10:31 AM
Amer
Quote:

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
• Jun 8th 2009, 11:35 AM
HallsofIvy
By the way, "proof by induction", which is what is meant here, and "inductive reasoning" are not at all the same thing!
• Jun 8th 2009, 09:22 PM
woof
Quote:

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:
$k(2k+1)+(4(k+1)-1)=2k^2+k+(4k+3)=2k^2+5k+3=(k+1)(2k+3)$