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Question

Let sum the following series to n terms and infinity $$\displaystyle \frac{1}{1.4.7} + \frac{1}{4.7.10} + \frac{1}{7.10.13} + ......$$ be $$\displaystyle \frac{1}{k} - \frac{1}{m(3n + 1)(3n + 4)}, \frac{1}{24}$$.Find $$k-m-l$$ ?


Solution

$$\displaystyle \frac{1}{1.4.7} + \frac{1}{4.7.10} + \frac{1}{7.10.13} + ......$$

$$T_n = \displaystyle \frac{1}{(3n - 2)(3n + 1)(3n + 4)}$$

$$=\displaystyle \frac{1}{3n + 1} \left ( \frac{1}{3n - 2} - \frac{1}{3n + 4} \right )\frac{1}{6}$$

$$=\displaystyle  \left ( \frac{1}{(3n - 2)(3n + 1)} - \frac{1}{(3n + 1)(3n + 4)} \right )\frac{1}{6}$$

$$S_n =\displaystyle  \frac{1}{6}\left [ \frac{1}{1.4} - \frac{1}{4.7} + \frac{1}{4.7} - \frac{1}{7.10} + \frac{1}{(3n + 1)(3n + 4)} \right ]$$

$$= \displaystyle \frac{1}{6} \left [ \frac{1}{4} - \frac{1}{(3n + 1)(3n + 4)} \right ]$$

Now $$S_\infty  = \dfrac{1}{24}$$

$$k=24,m=6,l=4$$

$$k-m-l=14$$

Maths

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