Crazy Minimum

This problem is mainly utilizing formula for sum of two squares.

$\displaystyle \sum_{x=1}^{k} x^2 = \dfrac{k(k+1)(2k+1)}{6}$

Using substitution, we get:

$\dfrac{\displaystyle \sum_{k=1}^{n} (2k)^2 }{\displaystyle \sum_{k=1}^{n} (2k-1)^2 }$

$\dfrac{\displaystyle 4 \sum_{k=1}^{n} k^{2} }{\displaystyle \sum_{k=1}^{n} \left( 4k^2 - 4k + 1 \right) }$

$\dfrac{4 \left[ \dfrac{(n)(n+1)(2n+1)}{6} \right] }{4 \left[ \dfrac{n(n+1)(2n+1)}{6} \right] - 4 \left[ \dfrac{n(n+1)}{2} \right] + n } < 1.01$

Factor out $n$

$\dfrac{4 \left[ \dfrac{(n+1)(2n+1)}{6} \right]}{\dfrac{4 (n+1)(2n+1)}{6} - \dfrac{4(n+1)}{2} + 1} < 1.01$

$\dfrac{4 \left[ \dfrac{(n+1)(2n+1)}{6}\right] }{\dfrac{4(n+1)(2n+1) - 3 \cdot 4(n+1) + 6}{6}} < 1.01$

$\dfrac{4 \left[ \dfrac{(n+1)(2n+1)}{6}\right] }{\dfrac{8n^{2} - 2}{6}} < 1.01$

$\dfrac{4 \left[ \dfrac{(n+1)(2n+1)}{6}\right] }{\dfrac{2 (2n+1)(2n-1) }{6}} < 1.01$

Now, mass canceling yields:

$\dfrac{2(n+1)}{2n-1} < 1.01$

Solving this linear inequality yields that $n>150.5$ , and the next integer that is greater than that (since this involves summation notation, we can only accept positive integers) is $\boxed{151}$

rewrite it as {100/101} < {summation squares of odd terms till 2n-1 / summation of squares of even term till 2n) . now interesting part comes here is to use componendo property so that in numerator we are left with summation of square till 2n tems and in denominator 4 x sum of squares of n terms . now use summation of square formula on solving we get simple equation as *(201/101)< [(4n+1)]/[2(2n+1)] * which results 1 50.5<n and thus ans is 151.