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you've got to take steps:
1) Show that P(1) is true
2) Assume P(k) is true
3) find P(k+1)
^_^

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?

theres a 1/4 lol

oh right >_< thank you :)

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

its the same as the other induction question you put essentually

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

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

except

are you getting the picture 2BGood? :)

ksajflkdaf

yep ^_^ , just following the steps dear

lol Mak :)

follow*

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

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

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

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

now its just a matter of algebra

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

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

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