The minimum value of n for which \frac{2^2+4^2+6^2+...+(2n)^2}{1^2+3^2+5^2+...+(2n-1)^2}

KVPY SB SX Exam 2011 Question Paper

Question : 82 of 120

Marks: +1, -0

The minimum value of n for which

{2^2+4^2+6^2+...+(2n)^2}/{1^2+3^2+5^2+...+(2n-1)^2} < 1.01

Solution:

x/{{2n(2n+1)(4n+1)}/6-x} < 1.01

2.01 x < (1.01){2n(2n+1)(4n+1)}/6

2.01 .{4n( n+1)(2n+1)}/6

< (1.01){2n(2n+1)(4n+1)}/6

2.01/1.01 < {4n+1}/{2n+2}⇒n > 150.5

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