1 + 1 3 x + 1.4 3.6 x^2 + 1.4.7 3.6.9 x^3 + .... is equal to - Vidyadip

MCQ

$1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + .... $ is equal to

A
$x$
B
${(1 + x)^{1/3}}$
C
${(1 - x)^{1/3}}$
✓
${(1 - x)^{ - 1/3}}$

✓

Answer

Correct option: D.

${(1 - x)^{ - 1/3}}$

d
(d) Let ${(1 + y)^n} = 1 + \frac{1}{3}x + \frac{{1.4}}{{3.6}}{x^2} + \frac{{1.4.7}}{{3.6.9}}{x^3} + ....$

$ = 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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