JEE Mains · Maths · STD 11 - 7. binomial theoram
The lowest integer which is greater than \(\left(1+\frac{1}{10^{100}}\right)^{10^{100}}\) is \(.....\)
- A \(3\)
- B \(4\)
- C \(2\)
- D \(1\)
Answer & Solution
Correct Answer
(A) \(3\)
Step-by-step Solution
Detailed explanation
Let \(P=\left(1+\frac{1}{10^{100}}\right)^{10^{100}}\), Let \(x=10^{100}\) \(\Rightarrow P=\left(1+\frac{1}{x}\right)^{x}\) \(\Rightarrow P=1+(x)\left(\frac{1}{x}\right)+\frac{(x)(x-1)}{\lfloor 2} \cdot \frac{1}{x^{2}}\)…
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