A Radical Recurrence

★ 7/10SubstitutionsequencesTelescoping

Problem

A sequence has a1=2a_1 = 2 and

an+1=19(an+24an+1+2).a_{n+1} = \frac19\left(a_n + 2\sqrt{4a_n + 1} + 2\right).

1. Find a formula for ana_n. 2. Let TnT_n be the sum of the first nn terms of {1an}\left\{\dfrac{1}{a_n}\right\}. Prove that

Tn>4n3−149.T_n > \frac{4n}{3} - \frac{14}{9}.

Answer

Solution

Difficulty7/10
TopicsSubstitution, sequences, Telescoping

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