Question

$1+\frac{1}{3} x+\frac{1.4}{3.6} x^{2}+\frac{1.4 .7}{3.6 .9} x^{3}+\cdots$ is equal to

• A. $x$
• B. $(1+x)^{1 / 3}$
• C. $(1-x)^{1 / 3}$
• D. $(1-x)^{-1 / 3}$

Answer 1

Answer: (D)

Let $(1+y)^{n}=1+\frac{1}{3} x+\frac{1 \cdot 4}{3 \cdot 6} x^{2}+\frac{1 \cdot 4 \cdot 7}{3 \cdot 6 \cdot 9} x^{3}+\cdots$
$=1+n y+\frac{n(n-1)}{2 !} y^{2}+\ldots$
Comparing the terms, we get
$n y=\frac{1}{3} x, \frac{n(n-1)}{2 !}$
$y^{2}=\frac{1 \cdot 4}{3 \cdot 6} x^{2}$
Solving, $n=-\frac{1}{3}, y=-x$
Hence, given series $=(1-x)^{-1 / 3}$.