Question

Please Help!

Answer 1

\[\frac{ k }{ 2 }\left( 3k + 1 \right) + \left( 3\left( k + 1\right) + 1 \right) = \frac{ \left( k + 1 \right) }{ 2 }\left( 3\left( k + 1 \right) +1 \right)\]

Answer 2

@Directrix
@mathslover
@jim_thompson5910
@.Sam.
@Callisto
@hartnn
@him1618
@wio

Answer 3

@ghazi

Answer 4

and what do you wish to accomplish here?

Answer 5

I need to prove they are equal.

Answer 6

It is proof by induction.

Answer 7

so, i would start by doing it out.
k/2(3k+1)+(3(k+1)+1)=(k+1)/2(3(k+1)+1)
k/2(3k+1)+(3k+3)+1)=(k+1)/2(3k+3)+1)
k/2(3k+1)+(3k+4)=(k+1)/2(3k+4)

Answer 8

... do you understand what i'm doing?

Answer 9

yep got that :)

Answer 10

@rosho

Answer 11

ok

Answer 12

@Directrix .?

Answer 13

i know but i need help through those steps.

Answer 14

actually i need assistance after what @rosho did and i had.

Answer 15

what rosho did was proving it normally, you have to prove it by induction.

Answer 16

by the way i have to use the Principle of Mathematical Induction and Show All Steps.

Answer 17

I see.

Answer 18

*show all 3 steps^

Answer 19

from http://www.math.toronto.edu/oz/turgor/Induction.pdf

Principle of Mathematical Induction:
If it is known that
(1) some statement is true for n = 1
(2) assumption that statement is true for n implies that the statement is true for
(n + 1)
then the statement is true for all positive integers

Answer 20

i anit eve learn this yet wow

Answer 21

so start with proving the statement is true for k=1

Answer 22

that is what i am doing. the equation i posted is already set up for that.

Answer 23

from there i just need to prove it.

Answer 24

so you have the instance k=1 written down and calculated out?

Answer 25

yeah

Answer 26

not calculated

Answer 27

though

Answer 28

calculate it out to prove that it is equal

Answer 29

yeah

Answer 30

then move on to the second step: proving that the statement being true for K also means that the statement is true for K+1

Answer 31

ok