A) 100
B) 1000
C) 10000
D) 100000
Correct Answer: C
Solution :
\[{{(1+100)}^{100}}=1+100.100+\frac{100.99}{1.2}.{{(100)}^{2}}\]\[+\frac{100.99.98}{1.2.3}{{(100)}^{3}}+....\] \[{{(101)}^{100}}-1=100.100\left[ 1+\frac{100.99}{1.2}+\frac{100.99.98}{1.2.3}.100+.... \right]\] From above it is clear that, \[{{(101)}^{100}}-1\]is divisible by (100)2 = 10000
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