If S_n = 1 + frac{1}{2} + frac{1}{2^2} + dots | vedclass

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(C) The given series is a geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.The sum of the first $n$ terms is given by

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$8$

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(C) The given series is a geometric progression with first term $a = 1$ and common ratio $r = \frac{1}{2}$.The sum of the first $n$ terms is given by $S_n = \frac{a(1 - r^n)}{1 - r} = \frac{1(1 - (1/2)^n)}{1 - 1/2} = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} = 2 - \frac{1}{2^{n-1}}$.We are given the condition $2 - S_n Substituting $S_n$,we get $2 - (2 - \frac{1}{2^{n-1}}) This implies $2^{n-1} > 100$.We know that $2^6 = 64$ and $2^7 = 128$.Therefore,$n - 1 \geq 7$,which gives $n \geq 8$.The least integral value of $n$ is $8$.

If $S_n = 1 + \frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{n-1}}$,then the least integral value of $n$ such that $2 - S_n < \frac{1}{100}$ is

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