STD 11 - 8. sequence and series — Maths STD 11 Science — Question

$S = 1 + 3 + 7 + 15 + 31 + ...... + {T_n}$

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 $n = 1,\;2$.