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

Algebra ->  Sequences-and-series -> 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      Log On


   

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:

                     1%2Fa%5Bn%2B1%5D-1%2Fa%5Bn%5D=2.

Express the general term an in terms of n when a1 = 1%2F6

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

                     b%5Bn%5D=sqrt%281%2Bn%5E2%29-n
                     
                     Prove that b%5Bn%5D%3E=b%5Bn%2B1%5D
Found 3 solutions by Edwin McCravy, AnlytcPhil, ikleyn:
Answer by Edwin McCravy(20056) About Me  (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 = 1%2F6

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 a%5B1%5D=1%2F6, 

matrix%281%2C5%2C1%5B%22%22%5D%2Fa%5Bn%2B1%5D%2C%22%22%2C%22%22=%22%22%2C%22%22%2C2n%2B6%29

Take reciprocals of both sides:

matrix%281%2C5%2Ca%5Bn%2B1%5D%2C%22%22%2C%22%22=%22%22%2C%22%22%2C1%2F%282n%2B6%29%29

Let n = m-1


matrix%281%2C5%2Ca%5Bm-1%2B1%5D%2C%22%22%2C%22%22=%22%22%2C%22%22%2C1%2F%282m-2%2B6%29%29
matrix%281%2C5%2Ca%5Bm%5D%2C%22%22%2C%22%22=%22%22%2C%22%22%2C1%2F%282m%2B4%29%29

Let m = n

matrix%281%2C5%2Ca%5Bn%5D%2C%22%22%2C%22%22=%22%22%2C%22%22%2C1%2F%282n%2B4%29%29

-----------------------------------------------------
Edwin

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

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

        b%5Bn%5D=sqrt%281%2Bn%5E2%29-n

         Prove that b%5Bn%5D%3E=b%5Bn%2B1%5D

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

b%5Bn%5D%3Cb%5Bn%2B1%5D

sqrt%281%2Bn%5E2%29-n%3Csqrt%281%2B%28n%2B1%29%5E2%29-%28n%2B1%29

sqrt%281%2Bn%5E2%29-n%3Csqrt%281%2B%28n%2B1%29%5E2%29-n-1

sqrt%281%2Bn%5E2%29%3Csqrt%281%2Bn%5E2%2B2n%2B1%29-1

sqrt%281%2Bn%5E2%29%3Csqrt%28n%5E2%2B2n%2B2%29-1

sqrt%281%2Bn%5E2%29%2B1%3Csqrt%28n%5E2%2B2n%2B2%29

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

%281%2Bn%5E2%29%2B2sqrt%281%2Bn%5E2%29%2B1%3Cn%5E2%2B2n%2B2

1%2Bn%5E2%2B2sqrt%281%2Bn%5E2%29%2B1%3Cn%5E2%2B2n%2B2

2%2Bn%5E2%2B2sqrt%281%2Bn%5E2%29%3Cn%5E2%2B2n%2B2

Isolate the radical term on the left:

2sqrt%281%2Bn%5E2%29%3C2n

Divide both sides by 2

sqrt%281%2Bn%5E2%29%3Cn

Square both sides

1%2Bn%5E2%3Cn%5E2

1%3C0

A contradiction, so the inequality is proved.

b%5Bn%5D%3E=b%5Bn%2B1%5D

Edwin

Answer by ikleyn(52782) About Me  (Show Source):
You can put this solution on YOUR website!
.
I will answer question (i) here. 

You are given the sequence a%5Bn%5D such that 1%2Fa%5Bn%2B1%5D - 1%2Fa%5Bn%5D = 2 and a%5B1%5D = 1%2F6.


Then consider the sequence c%5Bn%5D determined as c%5Bn%5D = 1%2Fa%5Bn%5D for all n = 1, 2, 3, . . . 


Then c%5B1%5D = 6 and c%5Bn%2B1%5D = c%5Bn%5D%2B2.


It is clear that c%5Bn%5D is nothing else as an arithmetic progression with the first term 6 and the common difference 2.


So, c%5Bn%5D = 6 + (n-1)*2 = 4 + 2n.


Then a%5Bn%5D = 1%2Fc%5Bn%5D = 1%2F%282n%2B4%29 for all n = 1, 2, 3, . . .

Solved.