Binomial Theorem And Its Simple Applications Question 68
Question: If the coefficient of x in the expansion of $ {{( x^{2}+\frac{k}{x} )}^{5}} $ is 270, then k =
[EAMCET 2002]
Options:
A) 1
B) 2
C) 3
D) 4
Show Answer
Answer:
Correct Answer: C
Solution:
- $ {T_{r+1}}={}^{5}C_{r}{{(x^{2})}^{5-r}}{{( \frac{k}{x} )}^{r}} $ For coefficient of x, $ 10-2r-r=1\Rightarrow r=3 $
Hence, $ {T_{3+1}}={}^{5}C_3{{(x^{2})}^{5-3}}{{( \frac{k}{x} )}^{3}} $ According to question, $ 10,k^{3}=270\Rightarrow k=3 $ .
BETA
