KVPY SA Exam 2012 Question Paper

Question : 5 of 80

Marks: +1, -0

For

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

to exceed 1.01, the maximum value of n is

Solution:

{2^2(1^2+2^2+3^2+....+n^2)}/{[1^2+2^2+3^2+...+(2n)^2]-2^2(1^2+2^2+3^2+...+n^2)}

={{2^2n(n+1)(2n+1)}/6}/{{2n(2n+1)(4n+1)}/6-{2^2n(n+1)(2n+1)}/6} > 1.0

={2n+2}/{2n-1} > 1.01

=1+3/{2n-1} > 1.01

2n – 1 > 300
n < 150.5

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