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Question Number 57617 by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19

Commented by tanmay.chaudhury50@gmail.com last updated on 08/Apr/19

$s o u r c e H a l l a n d k n i g h t a l g e b r a$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

	$28) \frac{a_{1}}{1 + a_{1}} = 1 - \frac{1}{1 + a_{1}}$ $\frac{a_{2}}{(1 + a_{1}) \times (1 + a_{2})} - \frac{1}{1 + a_{1}} = \frac{a_{2} - 1 - a_{2}}{(1 + a_{1}) (1 + a_{2})} = - \frac{1}{(1 + a_{1}) (1 + a_{2})}$ $\frac{a_{3}}{(1 + a_{1}) (1 + a_{2}) (1 + a_{3})} - \frac{1}{(1 + a_{1}) (1 + a_{2})} = - \frac{1}{(1 + a_{1}) (1 + a_{2}) (1 + a_{3})}$ $. . . .$ $. . . .$ $n o w a d d t h e m$ $s o r e q u i r e d s u m i s$ $S = 1 - \frac{1}{(1 + a_{1}) (1 + a_{2}) (1 + a_{3}) . . . . (1 + a_{n})}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

	$26) \frac{1}{1 + x} + \frac{1}{1 - x} = \frac{2}{1 - x^{2}}$ $\frac{2}{1 + x^{2}} + \frac{2}{1 - x^{2}} = \frac{4}{1 - x^{4}}$ $\frac{4}{1 + x^{4}} + \frac{4}{1 - x^{4}} = \frac{8}{1 - x^{8}}$ $. . . . .$ $. . . .$ $a d d t h e m$ $S + \frac{1}{1 - x} = \frac{2^{n}}{1 - x^{2^{n}}}$ $S = - \frac{1}{1 - x} + \frac{2^{n}}{1 - x^{2^{n}}}$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Apr/19

	$27)$ $\frac{x}{1 - x^{2}} + \frac{1}{1 - x^{2}} = \frac{x + 1}{1 - x^{2}} = \frac{1}{1 - x}$ $\frac{x^{2}}{1 - x^{4}} + \frac{1}{1 - x^{4}} = \frac{x^{2} + 1}{1 - x^{4}} = \frac{1}{1 - x^{2}}$ $t h a t m e a n s$ $\frac{x}{1 - x^{2}} = \frac{1}{1 - x} - \frac{1}{1 - x^{2}}$ $\frac{x^{2}}{1 - x^{4}} = \frac{1}{1 - x^{2}} - \frac{1}{1 - x^{4}}$ $. . .$ $. . .$ $\frac{x^{2^{n - 1}}}{1 - x^{2^{n}}} = \frac{1}{1 - x^{2^{n - 1}}} - \frac{1}{1 - x^{2^{n}}}$ $a d d t h e m$ $S = \frac{1}{1 - x} - \frac{1}{1 - x^{2^{n}}}$
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