SOLUTION: <pre>I need the answer for question (i) and (ii) both of them please: (2) Let {a<sub>n</sub>} and {b<sub>n</sub>} for n = 1,2,3,... . (i) The sequence {a<sub>n</sub>} satisfi

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Question 1085009: 
I need the answer for question (i) and (ii) both of them please:

(2) Let {an} and {bn} for n = 1,2,3,... .

(i) The sequence {an} satisfies the following relation:

                     .

Express the general term an in terms of n when a1 = 

(ii) The general term of {bn} is given by

                     
                     
                     Prove that

Found 3 solutions by Edwin McCravy, AnlytcPhil, ikleyn:
Answer by Edwin McCravy(20056)   (Show Source): You can put this solution on YOUR website!

(2) Let {an} and {bn} for n = 1,2,3,... .

(i) The sequence {an} satisfies the following relation:

                     .

Express the general term an in terms of n when a1 = 

Write down the the equation substituting n=1,2,3,4,...,n-1,n



Now add the equations and all the terms on the left cancel
except the first and last



So the sum of the left sides is the two terms that did not cancel,
and since there are n equations, the sum of the right sides is 2n. 





Since , 



Take reciprocals of both sides:



Let n = m-1





Let m = n



-----------------------------------------------------
Edwin
Answer by AnlytcPhil(1806)   (Show Source): You can put this solution on YOUR website!
Here's your second one:

(ii) The general term of {bn} is given by

        

         Prove that 

Assume that it is false for contradiction, that is we assume:













Since both sides are positive we can square both
sides and retain the inequality <







Isolate the radical term on the left:



Divide both sides by 2



Square both sides





A contradiction, so the inequality is proved.



Edwin
Answer by ikleyn(52785)   (Show Source): You can put this solution on YOUR website!
.
I will answer question (i) here. 

You are given the sequence  such that  -  = 2 and  = .


Then consider the sequence  determined as  =  for all n = 1, 2, 3, . . . 


Then  = 6 and  = .


It is clear that  is nothing else as an arithmetic progression with the first term 6 and the common difference 2.


So,  = 6 + (n-1)*2 = 4 + 2n.


Then  =  =  for all n = 1, 2, 3, . . .

Solved.



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