Let x be the mean of squares of first n natural numbers and y be the square of mean of first n natural numbers. If \frac{x}{y}=\frac{55}{42} then what is the value of n?

Explanation

We have x = \frac{\sum k^2}{n} = \frac{(n+1)(2n+1)}{6} and y = (\frac{\sum k}{n})^2 = \frac{(n+1)^2}{4}. Their ratio is \frac{x}{y} = \frac{(n+1)(2n+1)/6}{(n+1)^2/4} = \frac{2(2n+1)}{3(n+1)}. Setting this to \frac{55}{42} gives 84(2n+1) = 165(n+1), which simplifies to 168n + 84 = 165n + 165 \implies 3n = 81 \implies n = 27.

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