If p = 1/3 and q = 2/3, then what is the smallest odd n such that an + bn < 0.01?
Explanation:
For odd n, an + bn = p(n + 1)/2 q(n – 1)/2 + q(n + 1)/2 p(n – 1)/2 = p(n – 1)/2 q(n – 1)/2 (p + q)
Here, p = 1/3, q = 2/3
∴ p + q = 1
∴ an + bn = p(n – 1)/2 q(n – 1)/2 = (2/9)(n – 1)/2
Now considering options starting from the lowest,
For n = 7, an + bn = 8/729 ≈ 1/91 > 1/100
For n = 9, an + bn = 16/6561 ≈ 1/410 < 1/100
Hence, option (d).
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