Q56 / 90 · KVPY 2011 (SB/SX) · Mathematics

The minimum value of $\mathrm{n}$ for which $\frac{2^{2}+4^{2}+6^{2}+\ldots .+(2 \mathrm{n})^{2}}{1^{2}+3^{2}+5^{2}+\ldots .+(2 \mathrm{n}-1)^{2}} < 1.01$

  • A
    is 101
  • B
    is 121
  • Correct answer
    C
    is 151
  • D
    does not exist

Solution

$\frac{x}{\frac{2 n(2 n+1)(4 n+1)}{6}-x} < 1.01$ $2.01 \mathrm{x} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$ $2.01 \cdot \frac{4 n(n+1)(2 n+1)}{6} < (1.01) \frac{2 n(2 n+1)(4 n+1)}{6}$ $\frac{2.01}{1.01} < \frac{4 n+1}{2 n+2} \Rightarrow \mathrm{n}>150.5$

Hence, the correct answer is option: 'C'.

More Sequences and Series Questions