The sum of the series $\frac{3}{4 . 8} - \frac{3 . 5}{4 . 8 . 12} + \frac{3 . 5 . 7}{4 . 8 . 12 . 16} - ...$

• $\sqrt{\frac{3}{2}} - \frac{3}{4}$

• $\sqrt{\frac{2}{3}} - \frac{3}{4}$

• $\sqrt{\frac{3}{2}} - \frac{1}{4}$

• $\sqrt{\frac{2}{3}} - \frac{1}{4}$

$\frac{1}{2} - \frac{{\displaystyle 1}}{{\displaystyle 2 . 2^2}} + \frac{1}{3 . 2^3} - \frac{1}{4 . 2^4} + ...$ is equal to

• $\frac{1}{4}$

• ${\log }_{\mathrm{e}}\left(\frac{{\displaystyle 3}}{{\displaystyle 4}}\right)$

• ${\log }_{\mathrm{e}}\left(\frac{{\displaystyle 3}}{{\displaystyle 2}}\right)$

• ${\log }_{\mathrm{e}}\left(\frac{{\displaystyle 2}}{{\displaystyle 3}}\right)$

$\mathrm{For} \mathrm{any} \mathrm{integer} \mathrm{n} \ge 1, \mathrm{the} \mathrm{sum} \sum _{\mathrm{k} = 1}^{\mathrm{n}}\mathrm{k}\left(\mathrm{k} + 2\right) \mathrm{is} \mathrm{equal} \mathrm{to}$

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)\left(\mathrm{n} + 2\right)}{6}$

• $\frac{\mathrm{n}\left(\mathrm{n} +\hspace{0.17em}1\right)\left(2\mathrm{n} + 1\right)}{6}$

• $\frac{\mathrm{n}\left(\mathrm{n} + 1\right)\left(2\mathrm{n} + 7\right)}{6}$

• $\frac{\mathrm{n}\left(\mathrm{n} +\hspace{0.17em}1\right)\left(2\mathrm{n} + 9\right)}{6}$

$\mathrm{If} {\left(1 +\hspace{0.17em}\mathrm{x} + {\mathrm{x}}^2 + {\mathrm{x}}^3\right)}^5 = \sum _{\mathrm{k} = 0}^{15}{\mathrm{a}}_{\mathrm{k}}{\mathrm{x}}^{\mathrm{k}}, \mathrm{then} \underset{\mathrm{k} = 0}{\overset{7}{\sum {\mathrm{a}}_2}}\mathrm{k} = 0$

• 128

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• 512

• 1024

175.

$$\begin{gathered} \mathrm{If} \mathrm{\alpha } = \frac{5}{2 ! 3} + \frac{5 . 7}{3 ! 3^2} + \frac{5 . 7. 9}{4! 3^3} +\hspace{0.17em} . . . , \mathrm{then} \\ {\mathrm{\alpha }}^2 + 4\mathrm{\alpha } \mathrm{is} \mathrm{equal} \mathrm{to} \end{gathered}$$

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$\frac{1}{1 . 3} + \frac{{\displaystyle 1}}{{\displaystyle 2 . 5}} + \frac{{\displaystyle 1}}{{\displaystyle 3 . 7}} +\frac{{\displaystyle 1}}{{\displaystyle 4 . 9}} + . . . = ?$

• $2{\log }_{\mathrm{e}}\left(2\right) - 2$

• $2 - {\log }_{\mathrm{e}}\left(2\right)$

• $2{\log }_{\mathrm{e}}\left(4\right)$

• ${\log }_{\mathrm{e}}\left(4\right)$

If l, m, n are in arithmetic progression, then the straight line b + my + n = 0 will pass through the point

• (- 1, 2)

• (1, - 2)

• (1, 2)

• (2, 1)

$$\begin{gathered} \frac{1}{{\mathrm{e}}^{3\mathrm{x}}}\left({\mathrm{e}}^{\mathrm{x}} + {\mathrm{e}}^{5\mathrm{x}}\right) = {\mathrm{a}}_0 + {\mathrm{a}}_1\mathrm{x} + {\mathrm{a}}_2{\mathrm{x}}^2 + . . \\ \Rightarrow 2{\mathrm{a}}_1 + 2^3{\mathrm{a}}_3 + 2^5{\mathrm{a}}_5 + . . . = ? \end{gathered}$$

• e

• e ^{- 1}

• 1

• 0

Let n = 1! + 4! + 7! + . . . + 400!. Then ten's digit of n is

• 1

• 6

• 2

• 7

Let a = $\frac{{10}^{\mathrm{n}}}{\mathrm{n}!}$ for n = 1, 2, 3 . . . then the greatest  value of n for which a_n  is the greatest is

• 11

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• 8