Binomial Theorem And Its Simple Applications Question 29
Question: If $ y=3x+6x^{2}+10x^{3}+……..\infty $ , then $ \frac{1}{3}y-\frac{1.4}{3^{2}2}y^{2}+\frac{1.4.7}{3^{2}3}y^{3}-…..,\infty $ is equal to
Options:
A) x
B) $ 1-x $
C) $ 1 + x $
D) $ x^{x} $
Show Answer
Answer:
Correct Answer: A
Solution:
- [a] We have $ y=3x+6x^{2}+10x^{3}+…… $
$ \Rightarrow 1+y=1+3x+6x^{2}+10x^{3}+….. $
$ \Rightarrow 1+y={{(1-x)}^{-3}}\Rightarrow 1-x={{(1+y)}^{-1/3}} $
$ \Rightarrow x=1-{{(1+y)}^{-1/3}} $
$ =\frac{1}{3}y-\frac{1.4}{3^{2}.2}y^{2}+\frac{1.4.7}{3^{2}.3}y^{3}-…….. $
BETA
