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
✓
Answer
Correct option: B.
${2^{n + 1}} - n - 2$
b
(b) Let ${T_n}$ be the ${n^{th}}$ term and $S$ the sum upto $n$ terms.
(b) Let ${T_n}$ be the ${n^{th}}$ term and $S$ the sum upto $n$ terms.
$S = 1 + 3 + 7 + 15 + 31 + ...... + {T_n}$
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 $n = 1,\;2$.
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