Problem 1

Calculus Level 3

Given a sequence with the general term x n = 4 n − 5 9 n + 2 x_n = \dfrac{4n-5}{9n+2} . Let a = lim ⁡ n → ∞ x n \displaystyle a = \lim_{n\to\infty} x_n and let the number of terms x n x_n lying outside the open interval ( a − 0.1 , a + 0.1 ) (a-0.1, a +0.1) be z z . Find 2 z 2z .


The answer is 12.

Anupam Khandelwal by Anupam Khandelwal , 24, India

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2 solutions

Kenny Lau
Jul 11, 2014

lim ⁡ n → ∞ 4 n − 5 9 n + 2 = 4 9 \lim_{n\rightarrow\infty}\frac{4n-5}{9n+2}=\frac49

4 n − 5 9 n + 2 < 4 9 − 0.1 4 n − 5 9 n + 2 < 31 90 90 ( 4 n − 5 ) < 31 ( 9 n + 2 ) 360 n − 450 < 279 n + 62 81 n < 512 n < 6.7 \begin{array}{rcl} \frac{4n-5}{9n+2}&<&\frac49-0.1\\ \frac{4n-5}{9n+2}&<&\frac{31}{90}\\ 90(4n-5)&<&31(9n+2)\\ 360n-450&<&279n+62\\ 81n&<&512\\ n&<&6.7 \end{array}

4 n − 5 9 n + 2 > 4 9 + 0.1 4 n − 5 9 n + 2 > 49 90 90 ( 4 n − 5 ) > 49 ( 9 n + 2 ) 360 n − 450 > 441 n + 98 81 n < − 548 n < − 6.8 \begin{array}{rcl} \frac{4n-5}{9n+2}&>&\frac49+0.1\\ \frac{4n-5}{9n+2}&>&\frac{49}{90}\\ 90(4n-5)&>&49(9n+2)\\ 360n-450&>&441n+98\\ 81n&<&-548\\ n&<&-6.8 \end{array}

Therefore n<6.7, z=6, 2z=12.

1 Helpful 0 Interesting 0 Brilliant 0 Confused

Haha.. Same problem is posted twice... See this one

Prasun Biswas - 6 years, 6 months ago

By applying limits....we get a=4/9 Therefore... According to the range.. Only xi ( i= 1,2,3,4,5,6) has the value outside the given range

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