anonymous
  • anonymous
Use mathematical induction prove that: ¼+2*¼+4*¼+....+[2(n-1)+¼]=4n²-3n/4 these are sooo confusing
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
you've got to take steps: 1) Show that P(1) is true 2) Assume P(k) is true 3) find P(k+1) ^_^
anonymous
  • anonymous
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?
anonymous
  • anonymous
theres a 1/4 lol

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anonymous
  • anonymous
oh right >_< thank you :)
anonymous
  • anonymous
so after that, you do the calculation and if it's true proceed to step (2)
anonymous
  • anonymous
its the same as the other induction question you put essentually
anonymous
  • anonymous
expect this time it wont be so easy to simplify the expression in step 3
anonymous
  • anonymous
2) Assume p(k) :<-- in this case, replace the n's with k ^_^
anonymous
  • anonymous
except
anonymous
  • anonymous
are you getting the picture 2BGood? :)
anonymous
  • anonymous
i know it's like the other one, and you explained the other one so well, but i look at the problem and go????????????
anonymous
  • anonymous
ksajflkdaf
anonymous
  • anonymous
yep ^_^ , just following the steps dear
anonymous
  • anonymous
lol Mak :)
anonymous
  • anonymous
follow*
anonymous
  • anonymous
Assume S(k)= ( 4k^2 -3k) / 4 We need to prove that S(k+1) = [4(k+1)^2 -3(k+1) ] / 4
anonymous
  • anonymous
alright, elec got it from here, good luck :)
anonymous
  • anonymous
remember that I got the second line above by sub n=k+1 into the expression on the RHS of the sum
anonymous
  • anonymous
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
anonymous
  • anonymous
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
anonymous
  • anonymous
so we know that S(k+1) =[ [ 4k^2 -3k ] / 4] +( 2k + (1/4) )
anonymous
  • anonymous
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
anonymous
  • anonymous
now its just a matter of algebra
anonymous
  • anonymous
S(k+1) =[ 4k^2 -3k +8k +1 ] / 4 by getting over a common denominator
anonymous
  • anonymous
2BGood, are you getting the picture? ^_^
anonymous
  • anonymous
now , this is the exact same as what were asked to prove
anonymous
  • anonymous
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
anonymous
  • anonymous
all summmation induction questions are relatively straight forward, just S(k+1)= S(k) + T(k+1) , know thay and you are good
anonymous
  • anonymous
divisability induction questions are also relatively straight forward, but inequality induction questions are when it starts to get a bit tricky

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