Prove that
$\frac{C_{0}}{x} - \frac{C_{1}}{x + 1} + \frac{C_{2}}{x + 2} - (- 1)^{n} \frac{C_{n}}{x + n} = \frac{n!}{x (x + 1) . . (x + n)}$
where n is the + ive integer and x is not a negative integer Hence prove that
$\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{7} - . . . . . . . + (- 1)^{n} \frac{C_{n}}{3 n + 1} = \frac{3^{n} \cdot n!}{1.4.7..... (3 n + 1)}$

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Solution

Splitting the right hand side into partial functional by methods of suppression
i.e. by putting x = 0 , , -1 , -2 .. -n we get
$A_{0} = \frac{n!}{n!} = 1 = C_{0}$
$A_{1} = \frac{n!}{- 2.3 . . . . . (n - 1)} = \frac{- n!}{(n - 1)!} = - C_{1}$
$A_{2} = \frac{n!}{(- 1) (- 2.) 1.2.3 . . . . . (n - 2)} = \frac{- n!}{2! (n - 1)!} = C_{2}$
$A_{n} = (- 1)^{n} C_{n} .$
Hence we prove the first part , putting x = 1/3 in the first part we get second part .

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\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{7} - . . . . . . . + (- 1)^{n} \frac{C_{n}}{3 n + 1} = \frac{3^{n} \cdot n!}{1.4.7..... (3 n + 1)}

\frac{C_{0}}{1} - \frac{C_{1}}{4} + \frac{C_{2}}{7} - . . . + {(- 1)}^{n} \frac{C_{n}}{3 n + 1} =

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\frac{C_{0}}{x} - \frac{C_{1}}{x + 1} + \frac{C_{2}}{x + 2} - . . . . . . + {(- 1)}^{n} \frac{C_{n}}{x + n} =

C_{r}

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