Question

Use mathematical induction prove that:
¼+2*¼+4*¼+....+[2(n-1)+¼]=4n²-3n/4

these are sooo confusing

Answer 1

you've got to take steps:

1) Show that P(1) is true
2) Assume P(k) is true
3) find P(k+1)

^_^

Answer 2

now to show that p(1) is true :

P(1) : 2(1-1) = 4(1)^2-3(1)/4<-- are you sure of the equation?

Answer 3

theres a 1/4 lol

Answer 4

oh right >_< thank you :)

Answer 5

so after that, you do the calculation and if it's true proceed to step (2)

Answer 6

its the same as the other induction question you put essentually

Answer 7

expect this time it wont be so easy to simplify the expression in step 3

Answer 8

2) Assume p(k) :<-- in this case, replace the n's with k ^_^

Answer 9

except

Answer 10

are you getting the picture 2BGood? :)

Answer 11

i know it's like the other one, and you explained the other one so well, but i look at the problem and go????????????

Answer 12

ksajflkdaf

Answer 13

yep ^_^ , just following the steps dear

Answer 14

lol Mak :)

Answer 15

follow*

Answer 16

Assume S(k)= ( 4k^2 -3k) / 4

We need to prove that S(k+1) = [4(k+1)^2 -3(k+1) ] / 4

Answer 17

alright, elec got it from here, good luck :)

Answer 18

remember that I got the second line above by sub n=k+1 into the expression on the RHS of the sum

Answer 19

Now, we also know the result S(k+1) = S(k) +T(k+1) , very important result you need to know to summation induction questions

Answer 20

all you've got to do is plug in k+1 in the sequence then prove that the LHS and RHS are equal.

LHS= .... (simplify it)

then check if you get the same thing with the RHS >_< LOL

Answer 21

so we know that S(k+1) =[ [ 4k^2 -3k ] / 4] +( 2k + (1/4) )

Answer 22

the first fraction comes from the assumption , the second bracket comes from subbing n=k+1 into the general term on the LHS , thats our (k+1)th term

Answer 23

now its just a matter of algebra

Answer 24

S(k+1) =[ 4k^2 -3k +8k +1 ] / 4 by getting over a common denominator

Answer 25

2BGood, are you getting the picture? ^_^

Answer 26

now , this is the exact same as what were asked to prove

Answer 27

if you go up to the post where I made the assumption step , and expanded out the S(k+1) I wrote there, then you will get the same thing

Answer 28

all summmation induction questions are relatively straight forward, just S(k+1)= S(k) + T(k+1) , know thay and you are good

Answer 29

divisability induction questions are also relatively straight forward, but inequality induction questions are when it starts to get a bit tricky