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anonymous
 5 years ago
Use mathematical induction prove that:
¼+2*¼+4*¼+....+[2(n1)+¼]=4n²3n/4
these are sooo confusing
anonymous
 5 years ago
Use mathematical induction prove that: ¼+2*¼+4*¼+....+[2(n1)+¼]=4n²3n/4 these are sooo confusing

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you've got to take steps: 1) Show that P(1) is true 2) Assume P(k) is true 3) find P(k+1) ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now to show that p(1) is true : P(1) : 2(11) = 4(1)^23(1)/4< are you sure of the equation?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right >_< thank you :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so after that, you do the calculation and if it's true proceed to step (2)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its the same as the other induction question you put essentually

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0expect this time it wont be so easy to simplify the expression in step 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02) Assume p(k) :< in this case, replace the n's with k ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0are you getting the picture 2BGood? :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i know it's like the other one, and you explained the other one so well, but i look at the problem and go????????????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yep ^_^ , just following the steps dear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Assume S(k)= ( 4k^2 3k) / 4 We need to prove that S(k+1) = [4(k+1)^2 3(k+1) ] / 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0alright, elec got it from here, good luck :)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0remember that I got the second line above by sub n=k+1 into the expression on the RHS of the sum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Now, 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
 5 years ago
Best ResponseYou've already chosen the best response.0all 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
 5 years ago
Best ResponseYou've already chosen the best response.0so we know that S(k+1) =[ [ 4k^2 3k ] / 4] +( 2k + (1/4) )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the 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
 5 years ago
Best ResponseYou've already chosen the best response.0now its just a matter of algebra

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0S(k+1) =[ 4k^2 3k +8k +1 ] / 4 by getting over a common denominator

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02BGood, are you getting the picture? ^_^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0now , this is the exact same as what were asked to prove

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0all summmation induction questions are relatively straight forward, just S(k+1)= S(k) + T(k+1) , know thay and you are good

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0divisability induction questions are also relatively straight forward, but inequality induction questions are when it starts to get a bit tricky
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