Question

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

• A. is $101$
• B. is $121$
• C. is $151$
• D. does not exist

Answer 1

Answer: (C)

is $151$

$\sum \frac{4n^2}{{(2n-1)}^2}=\frac{4n(n+1)(2n+1)/6}{\sum (4n^2+1-4n)}\frac{4n(n+1)(2n+1)/6}{\frac{4n(n+1)(2n+1)}{6}+\frac{6n}{6}-\frac{12n(n+1)}{6}}<1.01\frac{4(n+1)(2n+1)}{4(n+1)(2n+1)+6-12(n+1)}<1.01\frac{4(n+1)(2n+1)}{(n+1)(8n+4-12)+6}<1.01\frac{4(n+1)(2n+1)}{(8n^2-8+6)}<1.01\frac{4(n+1)(2n+1)}{2(2n-1)(2n+1)}<1.01\frac{2(n+1)}{(2n-1)}<1.012n+2<20.2n-1.013.01<0.02n150.5<nn=151$