8. sequence and series — ગણિત ધોરણ 11 સાયન્સ — Question

$S_{n - 1}= 1 + 3 + 7 + 15 + …+ t_{n-1}$

હવે $,\text{ }{{\text{t}}_{\text{n}}}\text{ = }{{\text{S}}_{\text{n}}}\text{ - }{{\text{S}}_{\text{n-1}}}\text{ }$ $\text{= 1 + 2 + }{{\text{2}}^{\text{2}}}\text{ + }{{\text{2}}^{\text{3}}}\text{ + + n}$ પદ 

$\text{ }=\frac{2({{2}^{n}}-1)}{2-1}$

$S_n = 2^n - 1$  હવે, $S_n = 1 + 3 + 7 + 15 + 31 + ….+ n$ પદસુધી

$=\sum{{{t}_{n}}}\,\,\,=\sum{({{2}^{n}}-1)}\,\,\,=\sum{{{2}^{n}}}-\sum{1}\,\,$

$=(2+{{2}^{2}}+{{2}^{3}}+.....+{{2}^{n}})-n$   $=\frac{2({{2}^{n}}-1)}{2-1}-n\,\,$

$={{2}^{n+1}}-2-n$