MCQ
$1 + 3 + 7 + 15 + 31 + ..........$to $n$ terms =
- A${2^{n + 1}} - n$
- ✓${2^{n + 1}} - n - 2$
- C${2^n} - n - 2$
- DNone of these
$S = 1 + 3 + 7 + 15 + 31 + ...... + {T_n}$
Again $S = 1 + 3 + 7 + 15 + ...........{\rm{ }} + {T_{n - 1}} + {T_n}$
Subtracting, we get $0 = 1 + \left\{ {2 + 4 + 8 + ...({T_n} - {T_{n - 1}})} \right\} - {T_n}$
$\therefore \;\;{T_n} = 1 + 2 + {2^2} + {2^3} + .....{\rm{upto}}\;n\;{\rm{terms}}$
$ = \frac{{1({2^n} - 1)}}{{2 - 1}} = {2^n} - 1$
Now $S = \Sigma {T_n} = \Sigma {2^n} - \Sigma 1$
$ = (2 + {2^2} + {2^3} + ...... + {2^n}) - n$
$ = 2\left( {\frac{{{2^n} - 1}}{{2 - 1}}} \right) - n = {2^{n + 1}} - 2 - n$.
Aliter : $1 + 3 + 7 + ...... + {T_n}$
$ = 2 - 1 + {2^2} - 1 + {2^3} - 1 + .......... + {2^n} - 1$
$ = (2 + {2^2} + ...... + {2^n}) - n = {2^{n + 1}} - 2 - n$.
Trick : Check the options for $n = 1,\;2$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
$\mathrm{x}_{\mathrm{i}}$ $\ \ 3\ \ 8\ \ 11\ \ 10\ \ 5\ \ 4$
$\mathrm{f}_{\mathrm{i}}$ $\ \ 5 \ \ 2 \ \ 3 \ \ 2 \ \ 4 \ \ 4$
Match each entry in List-$I$ to the correct entries in List-$II$.
| List-$I$ | List-$II$ |
| ($P$) The mean of the above data is | $(1) 2.5$ |
| ($Q$) The median of the above data is | $(2) 5$ |
| ($R$) The mean deviation about the mean of the above data is | $(3) 6$ |
| ($S$) The mean deviation about the median of the above data is | $(4) 2.7$ |
| $(5) 2.4$ |
The correct option is :