MCQ
The greatest integer which divides the number ${101^{100}} - 1$, is
- A$100$
- B$1000$
- ✓$10000$
- D$100000$
${(101)^{100}} - 1 = 100.100\left[ {1 + \frac{{100.99}}{{1.2}} + \frac{{100.99.98}}{{1.2.3}}.100 + ....} \right]$
From above it is clear that,
${(101)^{100}} - 1$ is divisible by $(100)^2$ $= 10000$
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$\sin ^{-1}\left(\sum_{i=1}^{\infty} x^{i+1}-x \sum_{i=1}^{\infty}\left(\frac{x}{2}\right)^i\right)=\frac{\pi}{2}-\cos ^{-1}\left(\sum_{i=1}^{\infty}\left(-\frac{x}{2}\right)^i-\sum_{i=1}^{\infty}(-x)^i\right)$
lying in the interval $\left(-\frac{1}{2}, \frac{1}{2}\right)$ is. . . . .
(Here, the inverse trigonometric functions $\sin ^{-1} x$ and $\cos ^{-1} x$ assume values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ and $[0, \pi]$, respectively.)