A) \[2x\]
B) \[{{x}^{2}}\]
C) \[2{{x}^{3}}\]
D) All of these
Correct Answer: B
Solution :
\[{{(1+x)}^{n}}=1+nx+\frac{n\,(n-1)}{2!}{{x}^{2}}+\frac{n(n-1)\,(n-2)}{3!}{{x}^{3}}+.....\] Þ \[{{(1+x)}^{n}}-nx-1\] \[={{x}^{2}}\left[ \frac{n(n-1)}{2!}+\frac{n(n-1)(n-3)}{3!}x+..... \right]\] 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., \[{{x}^{2}}=9\].
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