Question

For a given statement Pn, write the statements P1, Pk, and Pk+1.

Answer 1

\[2+4+6+...+2n=n(n+1)\]

Answer 2

I have already figured out that n=1 is true and I do not know how to solve \[2+4+6+...+2k=k(k+1)\]

Answer 3

you get to assume that
\[2+4+6+...+2k=k(k1) \] and use that to prove that
\[2+4+6+...+2k+2(k+1)=(k+1)(k+2)\]

Answer 4

sorry i mean you get to assume that
\[2+4+6+...+2k=k(k+1)\]

Answer 5

Oh okay so you just change form to n=k, but you do not solve it?

Answer 6

replace the \(2+4+6+...+2k\) in the expression
\[2+4+6+...+2k+2(k+1)\] by \(k(k+1)\) "by induction"
then see if you can prove that
\[k(k+1)+2(k+1)=(k+1)(k+2)\] by algebra

Answer 7

wait where did the k(k+1) come from at the other side? I thought it would be
\[2(k+1)=k+1(k+1+1)\]

Answer 8

I'm just sitting here learning

Answer 9

ok lets go slow

Answer 10

have you ever done a proof by induction or is this the first?

Answer 11

you get to assume the statement \(P_k\) which is what you get when you replace \(n\) by \(k\)

Answer 12

clarifications on the inductive step "assume n=k"

\[2+4+6+\cdots+2k=k(k+1)\]
the left hand side consists of k terms
the right hand side is an expression equivalent to the sum of k terms

prove true for n=k+1
the left hand side has k+1 terms, \((2+4+6+\cdots+2k)+2(k+1)\)
the first k terms is enclosed in parentheses, the (k+1)st (added) term is \(2(k+1)\)

the right hand side is an expression obtained by replacing n by k+1
\[(k+1)([k+1]+1)\] or \[(k+1)(k+2)\] after simplification

Answer 13

in other words you get to assume that
\[2+4+6+...+2k=k(k+1)\]

Answer 14

whoa wait what? This is the first problem I have seen like this one, but I have done mathematical induction before

Answer 15

we take that as a given

Answer 16

in other words, we assume that we know that
\[2+4+6+...+2k=k(k+1)\]
is that part okay?

Answer 17

yes

Answer 18

now using the above, you are trying to show that the statement \(P_{k+1}\) is true
the statement \(P_{k+1}\) is what you get when you replace \(n\) by \(k+1\)

Answer 19

that is, you are going to try to show that
\[2+4+6+...+2k+2(k+1)=(k+1)(k+2)\]
is that part okay?

Answer 20

wait how did you get the right side?

Answer 21

i replaced every \(n\) in the original statement by \(k+1\)

Answer 22

oh okay so its k+1+1 which is k+2 I ssee that now

Answer 23

it was \(n(n+1)\) if you put \(n=k+1\) you get \((k+1)(k+2)\)

Answer 24

so far so good?

Answer 25

yes

Answer 26

now we use the "induction hypotheses" which means we get to replace \(2+4+6+...+2k\) by \(k(k+1)\)

Answer 27

that is called the "inductive step"
do you understand that? we use what we get to assume

Answer 28

\[\overbrace{2+4+6+...+2k}+2(k+1)\] that part on the left gets replaces by \(k(k+1)\)

Answer 29

okay so then its \[k(k+1)+k(k+1)=(k+1)(k+2)\]

Answer 30

that is what you are going to have to show by algebra, that the left hand side is equal to the right hand side

Answer 31

oh wait, there is a mistake in what you wrote

Answer 32

it should be
\[k(k+1)+2(k+1)=(k+1)(k+2)\]

Answer 33

oops I just saw that

Answer 34

\[\overbrace{2+4+6+...+2k}+2(k+1)\]
\[k(k+1)+2(k+1)\]

Answer 35

you want to show that this is \((k+1)(k+2)\)
that is the algebra part. that is just algebra
the main part is trying to figure out how to get the proof started,

Answer 36

the algebra part is usually easy
in this case you have
\[k(k+1)+2(k+1)\] both terms have a common factor of \(k+1)\) so you can factor it and get
\[(k+2)(k+1)\] in one steep

Answer 37

oh okay! thank you for the help!

Answer 38

yw