limn→∞(11.4+14.7+17.10+...+1(3n-2)(3n+1))=

Ltn→∞(11.4+14.7+17.10+...+1(3n-2)(3n+1)) =Ltn→∞131-14+14-17+17-110+……+13n-2-13n+1 =Ltn→∞131-13n+1 =13(1-0)=13

limn→∞(11.4+14.7+17.10+...+1(3n-2)(3n+1))=

Ltn→∞(11.4+14.7+17.10+...+1(3n-2)(3n+1)) =Ltn→∞131-14+14-17+17-110+……+13n-2-13n+1 =Ltn→∞131-13n+1 =13(1-0)=13

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$\lim_{n \to \infty} (\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . + \frac{1}{(3 n - 2) (3 n + 1)}) =$

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$\frac{1}{2}$

$\frac{1}{3}$

$\frac{1}{6}$

answer is C.

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Detailed Solution

$\underset{n \to \infty}{Lt} (\frac{1}{1.4} + \frac{1}{4.7} + \frac{1}{7.10} + . . . + \frac{1}{(3 n - 2) (3 n + 1)}) = \underset{n \to \infty}{Lt} \frac{1}{3} [1 - \frac{1}{4} + \frac{1}{4} - \frac{1}{7} + \frac{1}{7} - \frac{1}{10} + \dots \dots + \frac{1}{3 n - 2} - \frac{1}{3 n + 1}] = \underset{n \to \infty}{Lt} \frac{1}{3} [1 - \frac{1}{3 n + 1}] = \frac{1}{3} (1 - 0) = \frac{1}{3}$