If f is a positive function such that f(x+T)=f(x) ,(T>0),∀x∈R then limn→∞n(f(x+T)+2f(x+2T)+…..+nf(x+nT)f(x+T)+4f(x+4T)+…..+n2f(x+n2T))=

f(x+T)=f(x+2T)=…..=f(x+nT)=f(x) limn→∞nf(x)(1+2+3+….+n)f(x)(1+22+32+…+n2)=limn→∞n(n(n+1)2)n(n+1)(2n+1)6=32

If f is a positive function such that f(x+T)=f(x) ,(T>0),∀x∈R then limn→∞n(f(x+T)+2f(x+2T)+…..+nf(x+nT)f(x+T)+4f(x+4T)+…..+n2f(x+n2T))=

f(x+T)=f(x+2T)=…..=f(x+nT)=f(x) limn→∞nf(x)(1+2+3+….+n)f(x)(1+22+32+…+n2)=limn→∞n(n(n+1)2)n(n+1)(2n+1)6=32

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If f is a positive function such that $f (x + T) = f (x), (T > 0), \forall x \in R$ then $\lim_{n \to \infty} n (\frac{f (x + T) + 2 f (x + 2 T) + \dots . . + nf (x + nT)}{f (x + T) + 4 f (x + 4 T) + \dots . . + n^{2} f (x + n^{2} T)}) =$

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$\frac{2}{3}$

$\frac{3}{2}$

answer is C.

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Detailed Solution

$f (x + T) = f (x + 2 T) = \dots . . = f (x + nT) = f (x) \lim_{n \to \infty} \frac{nf (x) (1 + 2 + 3 + \dots . + n)}{f (x) (1 + 2^{2} + 3^{2} + \dots + n^{2})} = \lim_{n \to \infty} \frac{n (\frac{n (n + 1)}{2})}{\frac{n (n + 1) (2 n + 1)}{6}} = \frac{3}{2}$