Binomial Theorem And Its Simple Applications Question 153
Question: If the coefficient of $ x^{7} $ in $ {{( ax^{2}+\frac{1}{bx} )}^{11}} $ is equal to the coefficient of $ {x^{-7}} $ in $ {{( ax-\frac{1}{bx^{2}} )}^{11}} $ , then ab =
[MP PET 1999; AMU 2001; Pb. CET 2002; AIEEE 2005]
Options:
A) 1
B) 1/2
C) 2
D) 3
Show Answer
Answer:
Correct Answer: A
Solution:
- In the expansion of $ {{( ax^{2}+\frac{1}{bx} )}^{11}} $ , the general term is $ {T_{r+1}}={{,}^{11}}C_{r}{{(ax^{2})}^{11-r}}{{( \frac{1}{bx} )}^{r}}={{,}^{,11}}C_{r}{a^{11-r}}\frac{1}{b^{r}}{x^{22-3r}} $
For x7, we must have 22 - 3r = 7
therefore r = 5, and the coefficient of x7 = $ ^{11}C_5.{a^{11-5}}\frac{1}{b^{5}}={{,}^{11}}C_5\frac{a^{6}}{b^{5}} $
Similarly, in the expansion of $ {{( ax-\frac{1}{bx^{2}} )}^{11}}, $ the general term is $ {T_{r+1}}={{,}^{11}}C_{r}{{(-1)}^{r}}\frac{{a^{11-r}}}{b^{r}}.{x^{11-3r}} $
For x-7 we must have, 11 - 3r = -7
therefore r = 6, and the coefficient of $ {x^{-7}} $ is $ ^{11}C_6\frac{a^{5}}{b^{6}}={{,}^{11}}C_5\frac{a^{5}}{b^{6}} $ .
As given, $ ^{11}C_5\frac{a^{6}}{b^{5}}={{,}^{11}}C_5\frac{a^{5}}{b^{6}}\Rightarrow ab=1 $ .
BETA
