Question

\(lim_{n→∞} \frac1{2^n}(\frac1{\sqrt{1-1/2^n}} +\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+...+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}})\) is equal to

• A. \(\frac1{2}\)
• B. 1
• C. 2
• D. -2

Answer 1

The Correct Option is C

\(I =lim_{n→∞} \frac1{2^n}(\frac1{\sqrt{1-1/2^n}} +\frac1{\sqrt{1-\frac2{2^n}}}+\frac1{\sqrt{1-\frac3{2^n}}}+...+\frac1{\sqrt{1-\frac{2^n-1}{2^n}}})\) 

Let \(2^n=t \) and if\( n→∞\) then \(t→∞ \)

\(I= lim_{n→∞} \frac1{t} (\overset{t=1}{\underset{r=1}\sum} \frac1{√1-\frac{r}{t}})\)

\(l= ∫_0 ^1 \frac{dx}{√1-x}=∫_0^1 \frac {dx}{\sqrt {x}} \)      \(\bf{∫_0^a f(x)dx=∫_0^a f(a-x)dx }\)

\(=[2x^{\frac1{2}}]^1_0 \)

\(=2\)