A) by \[2x\]
B) by \[{{x}^{2}}\]
C) by \[2{{x}^{3}}\]
D) All of these
Correct Answer: B
Solution :
\[{{(1+x)}^{n}}=1+nx+\frac{n(n-1)}{1.2}{{x}^{2}}+\frac{n(n-1)(n-2)}{1.2.3}{{x}^{3}}+....\] \ \[{{(1+x)}^{n}}-nx-1={{x}^{2}}\left[ \frac{n(n-1)}{1.2}+\frac{n(n-1)(n-2)}{1.2.3}x+.... \right]\] From above it is clear that \[{{(1+x)}^{n}}-nx-1\]is divisible by x2.
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