If Sn-1-1-2-1-22-1-2n-1 then the least Value of n such that 2-Sn-1-100

The correct option is C 8 S_∞ = 2 #160; #160;S_∞ S_n lt; 1/100 #160;1/2^n + #160;1/2*2^n + ... lt; #160;1/100 2/2^n lt; ...

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Binomial Theorem

If Sn=1+1/...

Question

If $S_{n} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + . . . + \frac{1}{2^{n - 1}}$ then the least Value of n such that $2 - S_{n} < \frac{1}{100}$

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Solution

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Similar questions

S_{n} = 1 + \frac{1}{2} + \frac{1}{2^{2}} + \frac{1}{2^{3}} + . . . . . . . . . + \frac{1}{2^{n - 1}}

n

2 - S_{n} < \frac{1}{100}

S_{n} = [\frac{1}{1 + \sqrt{n}} + \frac{1}{2 + \sqrt{2 n}} + . . . + \frac{1}{n + \sqrt{n^{2}}}],

lim n \to \infty S_{n}

S_{n}

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . . . .,

S - S_{n} < \frac{1}{1000}

S

S_{n}

n

1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + . . .

S - S_{n} < \frac{1}{1000}

n

11

\sum_{r = 1}^{n} \frac{1 + 2 + 2^{2} + . . . Sum to r terms}{2^{r}}

1 - \frac{1}{2^{n}}

n - 1 + \frac{1}{2^{n}}

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