$\sum _{\mathrm{n} = 1}^{\infty }\frac{2{\mathrm{n}}^2 + \mathrm{n} + 1}{\mathrm{n}!}$ is equal to

• 2e - 1

• 2e + 1

• 6e - 1

• 6e + 1

If $\left|\mathrm{a}\right| <\hspace{0.17em}1, \mathrm{b} = \sum _{\mathrm{k} = 1}^{\infty }\frac{{\mathrm{a}}^{\mathrm{k}}}{\mathrm{k}}$, then a is equal to

• $\sum _{\mathrm{k} = 1}^{\infty }\frac{{\left(- 1\right)}^{\mathrm{k}}{\mathrm{b}}^{\mathrm{k}}}{\mathrm{k}}$

• $\sum _{\mathrm{k} = 1}^{\infty }\frac{{\left(- 1\right)}^{\mathrm{k} - 1}{\mathrm{b}}^{\mathrm{k}}}{\mathrm{k}!}$

• $\sum _{\mathrm{k} = 1}^{\infty }\frac{{\left(- 1\right)}^{\mathrm{k}}{\mathrm{b}}^{\mathrm{k}}}{\left(\mathrm{k} - 1\right)!}$

• $\sum _{\mathrm{k} = 1}^{\infty }\frac{{\left(- 1\right)}^{\mathrm{k} - 1}{\mathrm{b}}^{\mathrm{k}}}{\left(\mathrm{k} + 1\right)!}$

163.

$1 + \frac{2}{4} + \frac{2}{4} \frac{5}{8} + \frac{2}{4} \frac{5}{8} \frac{8}{12} + \frac{2}{4} \frac{5}{8} \frac{8}{12} \frac{11}{16} + . . . . \mathrm{is} \mathrm{equal} \mathrm{to}:$

• $4^{ - \raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

• $\sqrt[3]{16}$

• $\sqrt[3]{4}$

• $4^{\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}$

$\mathrm{If} \left|\mathrm{x}\right| < 1 \mathrm{and} \mathrm{y} = \mathrm{x} - \frac{{\mathrm{x}}^2}{2} + \frac{{\mathrm{x}}^3}{3} - \frac{{\mathrm{x}}^4}{4} + . . . , \mathrm{then} \mathrm{x} \mathrm{is} \mathrm{equal} \mathrm{to}$

• $\mathrm{y} + \frac{{\mathrm{y}}^2}{2} + \frac{{\mathrm{y}}^3}{3} + . . .$

• $\mathrm{y} - \frac{{\mathrm{y}}^2}{2} + \frac{{\mathrm{y}}^3}{3} - \frac{{\mathrm{y}}^4}{4} + . . .$

• $\mathrm{y} + \frac{{\mathrm{y}}^2}{2!} + \frac{{\mathrm{y}}^3}{3!} + . . .$

• $\mathrm{y} - \frac{{\mathrm{y}}^2}{2!} + \frac{{\mathrm{y}}^3}{3!} - \frac{{\mathrm{y}}^4}{4!} + . . .$

$\mathrm{If} {\mathrm{S}}_{\mathrm{n}} = 1^3 +\hspace{0.17em}{2}^3 + ... +\hspace{0.17em}{\mathrm{n}}^3 \mathrm{and} {\mathrm{T}}_{\mathrm{n}} = 1 +\hspace{0.17em}2 + ... + \mathrm{n}, \mathrm{then}$

• ${\mathrm{S}}_{\mathrm{n}} = {\mathrm{T}}_{{\mathrm{n}}^3}$

• ${\mathrm{S}}_{\mathrm{n}} = {\mathrm{T}}_{{\mathrm{n}}^2}$

• ${\mathrm{S}}_{\mathrm{n}} = {{\mathrm{T}}_{\mathrm{n}}}^2$

• ${\mathrm{S}}_{\mathrm{n}} = {{\mathrm{T}}_{\mathrm{n}}}^3$

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

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

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

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

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

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

• 128

• 256

• 512

• 1024

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

• 21

• 23

• 25

• 27

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

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

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

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

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

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)