Binomial Theorem And Its Simple Applications Question 19
Question: The term independent of x in the expansion of $ {{[({t^{-1}}-1)x+{{({t^{-1}}+1)}^{-1}}{x^{-1}}]}^{8}} $ is
Options:
A) $ 56{{( \frac{1-t}{1+t} )}^{3}} $
B) $ 56{{( \frac{1+t}{1-t} )}^{3}} $
C) $ 70{{( \frac{1-t}{1+t} )}^{4}} $
D) $ 70{{( \frac{1+t}{1-t} )}^{4}} $
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Answer:
Correct Answer: C
Solution:
- [c] $ [({t^{-1}}-1)x+{{({t^{-1}}+1)}^{-1}}{x^{-1}}){{]}^{8}} $
$ ={{[ ( \frac{1}{t}-1 )x+{{( \frac{1}{t}+1 )}^{-1}}\frac{1}{x} ]}^{8}} $
Let $ {T_{r+1}} $ be the term independent of x, then
$ {T_{r+1}}={{,}^{8}}C_{r}{{( \frac{1}{t}-1 )}^{8-r}}{x^{8-r}}\cdot {{( \frac{1}{t}+1 )}^{-r}}{{( \frac{1}{x} )}^{r}} $
$ ={{,}^{8}}C_{r}{{( \frac{1}{t}-1 )}^{8-r}}\cdot {{( \frac{1}{t}+1 )}^{-r}}\cdot {x^{8-2r}} $
$ \therefore 8-2r=0\Rightarrow r=0 $
$ \therefore T_5 $ is the term independent of x and $ T_5={{,}^{8}}C_4{{( \frac{1}{t}-1 )}^{4}}\cdot {{( \frac{1}{t}+1 )}^{-4}} $
$ ={{,}^{8}}C_4{{( \frac{1-t}{t} )}^{4}}\cdot {{( \frac{1+t}{t} )}^{-4}} $
$ ={{,}^{8}}C_4{{( \frac{1-t}{1+t} )}^{4}}=\frac{8.7.6.5}{4.3.2.1}{{( \frac{1-t}{1+t} )}^{4}} $
$ =,70.{{( \frac{1-t}{1+t} )}^{4}} $
BETA
