Binomial Theorem And Its Simple Applications Question 24
Question: $ 1+\frac{1}{3}x+\frac{1\times 4}{3\times 6}x^{2}+\frac{1\times 4\times 7}{3\times 6\times 9}x^{3}+… $ is equal to
Options:
A) x
B) $ {{(1+x)}^{1/3}} $
C) $ {{(1-x)}^{1/3}} $
D) $ {{(1-x)}^{-1/3}} $
Show Answer
Answer:
Correct Answer: D
Solution:
- [d] Let $ {{(1+y)}^{n}}=1+\frac{1}{3}x+\frac{1\times 4}{3\times 6}x^{2}+\frac{1\times 4\times 7}{3\times 6\times 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\times 4}{3\times 6}x^{2} $
Solving $ n=-1/3,y=-x $
Hence, the given series is $ {{(1-x)}^{-1/3}} $
BETA
