Question 1085009
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I will answer question (i) here.


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You are given the sequence {{{a[n]}}} such that {{{1/a[n+1]}}} - {{{1/a[n]}}} = 2 and {{{a[1]}}} = {{{1/6}}}.


Then consider the sequence {{{c[n]}}} determined as {{{c[n]}}} = {{{1/a[n]}}} for all n = 1, 2, 3, . . . 


Then {{{c[1]}}} = 6 and {{{c[n+1]}}} = {{{c[n]+2}}}.


It is clear that {{{c[n]}}} is nothing else as an arithmetic progression with the first term 6 and the common difference 2.


So, {{{c[n]}}} = 6 + (n-1)*2 = 4 + 2n.


Then {{{a[n]}}} = {{{1/c[n]}}} = {{{1/(2n+4)}}} for all n = 1, 2, 3, . . . 
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Solved.