Question

Please help... will fan and medal :)
For the given statement Pn, write the statements P1, Pk, and Pk+1.
(2 points)

2 + 4 + 6 + . . . + 2n = n(n+1)

Answer 1

whats the issue?

Answer 2

im not sure if im going to do the problem right and just need assistance

Answer 3

well, show me what you think should happen and ill correct as needed

Answer 4

P1 = 1(1+1)

Answer 5

is that right?

Answer 6

Pk= k(k+1)

Answer 7

Pk+1 = k+1(k+1+1) ??

Answer 8

thats only partly right

P1 = 2(1) = 1(1+1)

Answer 9

Pk is just Pn rewritten with k instead of n

Answer 10

Yeah i know that much im just not sure if i was doing it right for this specific problem.. i think im over thinking it ?

Answer 11

Pk = 2 + 4 + 6 + . . . + 2k = k(k+1)

Answer 12

P(k+1) = 2 + 4 + 6 + . . . + 2k + 2(k+1) = k(k+1) + 2(k+1)
| addon| | addon|

Answer 13

it looks like you might be starting proofs by induction?

Answer 14

yeah ..

Answer 15

Am i on the right track though ?

Answer 16

then it important to understand that:

P(k+1) = P(k) + (addon)

Answer 17

your dont fine so far, the idea is that we want to convert P(k)+(addon) into the same form as P(k)

in other words, we know the form of P(k) is good therefore its the form we want to establish

P(k) = k(k+1)

P(k+1) = P(k) + (addon) = k(k+1)+ (addon)

and then algebra the right side to the form (k+1)(k+1+1)

Answer 18

*** doing fine ....

Answer 19

idk what addon is :x

Answer 20

its the 'rule' for the end term in P(n)

in this case if 2(k+1) since the nth term is defined as 2n

Answer 21

2 + 4 + 6 + . . . + 2n
^^^
rule for the nth term


P(k) = 2 + 4 + 6 + . . . + 2k

P(k+1) = [2 + 4 + 6 + . . . + 2k] + [2(k+1)]

P(k+1) = P(k) + 2(k+1)

Answer 22

And thats it then?

Answer 23

thats what its going for yes

now remember that what you add to one side of an equals sign, you also add to the other

\[P(k)= 2 + 4 + 6 + . . . + 2k = k(k+1)\]
\[P(k+1)= \underbrace{\color{red}{2 + 4 + 6 + . . . + 2k}}_{P(k)}+2(k+1) = \underbrace{\color{red}{k(k+1)}}_{P(k)}+2(k+1)\]

Answer 24

we can now factor the right side since (k+1) is common to both terms

\[k(\color{green}{k+1})+2(\color{green}{k+1})\implies (\color{green}{k+1})(k+2)\]

Answer 25

I think im confusing myself lol im sorry ..

Answer 26

well,this is just a preview of the thought process you have to work with when doing induction proofs. just trying to get you exposed to it now so that its not such a shock in a few days :)

Answer 27

Oh .. im doing online schooling and this is the last question of this assignment .. and idk this one i just got stumped on it .. its from a previous lesson

Answer 28

Just to be clear.. this is not correct?
So for the answer i would then write :
p1= 2(1)=1(1+1)
Pk= 2k=k(k+1)
pK+1= 2(k+1)=k+1(k+1+1)

Answer 29

\[P(1):~2(1)=1(1+1)\]
\[P(k):~2+4+6+...+2k=k(k+1)\]
\[P(k+1):~2+4+6+...+2k+2(k+1)=k(k+1)+2(k+1)\\ \hspace{4em}\color{red}{\implies}(k+1)(k+1+1)\]

Answer 30

P(n) is not defined to be the last term .. its the sum of all the terms

Answer 31

gotchaaa .. thank you for that (:

Answer 32

P(1) = 2(1)

P(2) = 2(1) + 2(2)

P(3) = 2(1) + 2(2) + 2(3)

P(n) = 2(1) + 2(2) + 2(3) + ... + 2(n)

P(k) is not 2k

Answer 33

good :)

Answer 34

OHHHHHHHH okay .. that makes since.. i didnt realize thats how i was writing it.. thank you so much again

Answer 35

sense*