MCQ
$\mathop {\lim }\limits_{n \to \infty } \frac{1}{2} + \frac{1}{{{2^2}}} + \frac{1}{{{2^3}}} + ... + \frac{1}{{{2^n}}}$ equals
- A$2$
- B$-1$
- ✓$1$
- D$3$
$\mathop {\lim }\limits_{n \to \infty } \,\left[ {1 - \frac{1}{{{2^n}}}} \right] = 1 - 0 = 1$
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$f(n)=\frac{\sum_{k=0}^n \sin \left(\frac{k+1}{n+2} \pi\right) \sin \left(\frac{k+2}{n+2} \pi\right)}{\sum_{k=0}^n \sin ^2\left(\frac{k+1}{n+2} \pi\right)}$
Assuming $\cos ^{-1} x$ takes values in $[0, \pi]$, which of the following options is/are correct ?
$(1)$ $\sin \left(7 \cos ^{-1} f(5)\right)=0$
$(2)$ $f(4)=\frac{\sqrt{3}}{2}$
$(3)$ $\lim _{n \rightarrow \infty} f(n)=\frac{1}{2}$
$(4)$ If $\alpha=\tan \left(\cos ^{-1} f(6)\right)$, then $\alpha^2+2 \alpha-1=0$