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Study Stuff for MCA Examinations - NIMCET
Study Stuff for MCA Examinations - NIMCET
Previous Year Question (PYQs)
Let $[ \cdot ]$ denote the greatest integer function and
$f(x) = \lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right]$. Then $12\sum_{j=1}^{\infty} f(j)$ is equal to:
Solution
$\sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] \le \sum_{k=1}^{n} \frac{k^2}{3^x}$
$= \frac{n(n+1)(2n+1)}{6\cdot 3^x}$
$\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left[\frac{k^2}{3^x}\right] = \frac{1}{3^{x+1}}$
$\Rightarrow f(x) = \frac{1}{3^{x+1}}$
$12\sum_{j=1}^{\infty} f(j) = 12\sum_{j=1}^{\infty} \frac{1}{3^{j+1}}$
$= 12\left(\frac{1}{9} + \frac{1}{27} + \cdots\right)$
$= 12\left(\frac{1/9}{1 - 1/3}\right) = 2$
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