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If an = 3/4 - (3/4)2 + (3/2)2 + …….. (-1)n-1 (3/4)n and
bn = 1 - an, then find the least natural number n0 such that bn > an ∀ n ≥ no.

Simran Bhatia , 10 Years ago
Grade 11
anser 1 Answers
Aditi Chauhan

Last Activity: 10 Years ago

Hello Student,
Please find the answer to your question
an = 3/4 –(3/4)2 + (3/4)3 + …… + (-1)n-1 (3/4)n
= \frac{\frac{3}{4}(1-(\frac{-3}{4})^n)}{1+\frac{3}{4}}= 3/7 (1 – (-3/4)n)
Bn = 1 – an­ and bn > an ∀ n ≥ no
∴ 1 – an> an ⇒ 2an < 1
⇒ \frac{6}{7}[1 –(\frac{-3}{4})n] < 1 ⇒ - ( \frac{-3}{4})n < 1/6
⇒ (-3)n+1 < 22n-1
For n to be even, inequality always holds. For n to be odd, it holds for n ≥ 7.
∴ The least natural no, for which it holds is 6
( ∵ it holds for every even natural no.)
Thanks
Aditi Chauhan
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