1-2-1-4-1-8-1-2-n-1-1-2-n-P-r-o-v-e-the-above-for-integral-n-1-

Question Number 5723 by Rasheed Soomro last updated on 25/May/16

\frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots . + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}

P r o v e the above for integral n ⩾ 1 .

Commented by FilupSmith last updated on 25/May/16

LHS = S = \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \dots + \frac{1}{2^{n}}

S = \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n}}

2 S = 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n - 1}}

2 S + \frac{1}{2^{n}} = 1 + \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n - 1}} + \frac{1}{2^{n}}

2 S + \frac{1}{2^{n}} = 1 + S

∴ S = 1 - \frac{1}{2^{n}}

∴ \frac{1}{2} + \frac{1}{4} + \dots + \frac{1}{2^{n}} = 1 - \frac{1}{2^{n}}

Commented by Rasheed Soomro last updated on 25/May/16

G o o D A p p r o a c h!