Binomial Theorem And Its Simple Applications Question 26
Question: The coefficient of the term independent of x in the expansion of $ (1+x+2x^{3}){{( \frac{3}{2}x^{2}-\frac{1}{3x} )}^{9}} $ is
[DCE 1994]
Options:
A) $ \frac{1}{3} $
B) $ \frac{19}{54} $
C) $ \frac{17}{54} $
D) $ \frac{1}{4} $
Show Answer
Answer:
Correct Answer: C
Solution:
- The general term in the expansion of $ {{( \frac{3}{2}x^{2}-\frac{1}{3x} )}^{9}} $ is $ {T_{r+1}}={{,}^{9}}C_{r}{{( \frac{3}{2}x^{2} )}^{9-r}}{{( -\frac{1}{3x} )}^{r}} $
$ ={{,}^{9}}C_{r}{{( \frac{3}{2} )}^{9-r}}( -\frac{1}{3} ){x^{18-3r}} $ ……(i)
Now, the coefficient of the term independent of x in the expansion of $ (1+x+2x^{3}){{( \frac{3}{2}x^{2}-\frac{1}{3x} )}^{9}} $ ……(ii) = Sum of the coefficient of the terms $ x^{0},{x^{-1}} $ and $ {x^{-3}} $ in $ {{( \frac{3}{2}x^{2}-\frac{1}{3x} )}^{9}} $ . For $ x^{0} $ in (i) above, $ 18-3r=0\Rightarrow r=6 $ . For $ {x^{-1}} $ in (i) above, there exists no value of r and
Hence no such term exists.
For $ {x^{-3}} $ in (i), $ 18-3r=-3\Rightarrow r=7 $
$ \therefore $ For term independent of x, in (ii) the coefficient $ =1\times {{,}^{9}}C_6{{(-1)}^{6}}{{( \frac{3}{2} )}^{9-6}}{{( \frac{1}{3} )}^{6}}+2\times {{,}^{9}}C_7{{(-1)}^{7}}{{( \frac{3}{2} )}^{9-7}}{{( \frac{1}{3} )}^{7}} $
$ =\frac{9.8.7}{1.2.3}.\frac{3^{3}}{2^{3}}.\frac{1}{3^{6}}+2\frac{9.8}{1.2}(-1)\frac{3^{2}}{2^{2}}.\frac{1}{3^{7}} $
$ =\frac{7}{18}-\frac{2}{27}=\frac{17}{54} $ .
BETA
