An Alternating Coefficient Recursion

Problem

Sequences {an}\{a_n\} and {bn}\{b_n\} satisfy

bnan+an+1+bn+1an+2=0,bn=3+(1)n2,b_na_n + a_{n+1} + b_{n+1}a_{n+2} = 0, \qquad b_n = \frac{3 + (-1)^n}{2},

with a1=2a_1 = 2 and a2=4a_2 = 4.

1. Find a3,a4,a5a_3, a_4, a_5. 2. Let cn=a2n1+a2n+1c_n = a_{2n-1} + a_{2n+1}. Prove that {cn}\{c_n\} is geometric. 3. Let Sk=a2+a4++a2kS_k = a_2 + a_4 + \cdots + a_{2k}. Prove that

k=14nSkak<76(nN).\sum_{k=1}^{4n} \frac{S_k}{a_k} < \frac{7}{6} \qquad (n \in \mathbb{N}^*).

Answer

Solution

Difficulty8/10
TopicsGeometric Progression, Recursion, sequences, Telescoping, inequality

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