Binomial Theorem And Its Simple Applications Question 12
Question: In the binomial expansion $ {{(a+bx)}^{-3}} $ $ =\frac{1}{8}+\frac{9}{8}x+…. $ , then the value of a and b are:
Options:
A) a = 2, b = 3
B) a = 2, b = -6
C) a = 3, b = 2
D) a = -3, b = 2
Show Answer
Answer:
Correct Answer: B
Solution:
- [b] Given expansion is $ {{(a+bx)}^{-3}} $ which can be written as $ {{[ a( 1+\frac{b}{a}x ) ]}^{-3}}={a^{-3}}{{( 1+\frac{b}{a}x )}^{-3}} $
$ ={a^{-3}}( 1-\frac{3b}{a}x+6{{( \frac{b}{a}x )}^{2}}-…… ) $
$ (Byusing{{(1+x)}^{-3}}=1-3x+6x^{2}-………) $
But given that: $ {{(a+bx)}^{-3}}=\frac{1}{8}+\frac{9}{8}x+……. $
$ \therefore {a^{-3}}[ 1-\frac{3b}{a}x+6\frac{b^{2}}{a^{2}}.x^{2}-…. ]=\frac{1}{8}+\frac{9}{8}x+…. $
$ \Rightarrow ,{a^{-3}}=\frac{1}{8}={2^{-3}}\Rightarrow a=2 $
and $ -3b{a^{-4}}={{9.2}^{-3}}\Rightarrow b=-6 $
BETA
