MCQ
$1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + ....$ = . ..
- A$x$
- B${(1 + x)^{1/3}}$
- C${(1 - x)^{1/3}}$
- ✓${(1 - x)^{ - 1/3}}$
$ = 1 + ny + \frac{{n(n - 1)}}{{2!}}{y^2} + .....$
Comparing the terms, we get
$ny = \frac{1}{3}x,\frac{{n(n - 1)}}{{2!}}{y^2} = \frac{{1.4}}{{3.6}}{x^2}$
Solving, $n = - \frac{1}{3},y = - x$.
Hence given series $ = {(1 - x)^{ - 1/3}}$
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( જ્યાં $l$ એ સાન્ત સંખ્યા છે )