If f is periodic function with period T and [.] denotes greatest integer function thenLtn→∞⁡f(x+T)f(x+T)+22f(x+2T)f(x+2T)+32f(x+3T)f(x+3T)+…..+n2f(x+nT)f(x+nT)n3

Since f is a periodic function then f(x+T)=f(x+2T)=f(x+3T)=…=f(x+nT)=f(x)Ltn→∞⁡f(x)f(x)12+22+32+…..n2n3=Ltn→∞f(x)f(x) Ltn→∞∑ n2n3=f(x)f(x) Ltn→∞n(n+1)(2n+1)6n3=f(x)f(x) 3

If f is periodic function with period T and [.] denotes greatest integer function thenLtn→∞⁡f(x+T)f(x+T)+22f(x+2T)f(x+2T)+32f(x+3T)f(x+3T)+…..+n2f(x+nT)f(x+nT)n3

Since f is a periodic function then f(x+T)=f(x+2T)=f(x+3T)=…=f(x+nT)=f(x)Ltn→∞⁡f(x)f(x)12+22+32+…..n2n3=Ltn→∞f(x)f(x) Ltn→∞∑ n2n3=f(x)f(x) Ltn→∞n(n+1)(2n+1)6n3=f(x)f(x) 3

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If f is periodic function with period T and [.] denotes greatest integer function then
${Lt}_{n \to \infty} \frac{[f (x + T)^{f (x + T)}] + [2^{2} f (x + 2 T)^{f (x + 2 T)}] + [3^{2} f (x + 3 T)^{f (x + 3 T)}] + \dots . . + [n^{2} f (x + nT)^{f (x + nT)}]}{n^{3}}$

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1/3

$\frac{(f (x))^{f (x)}}{3}$

$\frac{(f (x))^{f (x)}}{4}$

$\frac{(f (x))^{f (x)}}{2}$

answer is A.

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Detailed Solution

$\begin{array}{l} Since f is a periodic function then f (x + T) = f (x + 2 T) = f (x + 3 T) = \dots = f (x + nT) = f (x) \\ {Lt}_{n \to \infty} \frac{[f (x)^{f (x)}] \{[1^{2}] + [2^{2}] + [3^{2}] + \dots . . [n^{2}]\}}{n^{3}} \\ = {Lt}_{n \to \infty} [f (x)^{f (x)}] {Lt}_{n \to \infty} \frac{\sum_{} n^{2}}{n^{3}} \\ = (f (x)^{f (x)} {Lt}_{n \to \infty} \frac{n (n + 1) (2 n + 1)}{6 n^{3}} \\ = \frac{(f (x)^{f (x)}}{3} \end{array}$