Binomial Theorem And Its Simple Applications Question 116
Question: $ {{(1+x)}^{n}}-nx-1 $ is divisible by (where $ n\in N $ )
Options:
A) $ 2x $
B) $ x^{2} $
C) $ 2x^{3} $
D) All of these
Show Answer
Answer:
Correct Answer: B
Solution:
- $ {{(1+x)}^{n}}=1+nx+\frac{n,(n-1)}{2!}x^{2}+\frac{n(n-1),(n-2)}{3!}x^{3}+….. $
therefore $ {{(1+x)}^{n}}-nx-1 $
$ =x^{2}[ \frac{n(n-1)}{2!}+\frac{n(n-1)(n-3)}{3!}x+….. ] $ From above it is clear that $ {{(1+x)}^{n}}-nx-1 $ is divisible by $ x^{2} $ . Trick: $ {{(1+x)}^{n}}-nx-1 $ . Put $ n=2 $ and $ x=3 $ ; Then $ 4^{2}-2.3-1=9 $ is not divisible by 6, 54 but divisible by 9. Which is given by option (b) i.e., $ x^{2}=9 $ .
BETA
