If `S_(n) = 1 + (1)/(2) + (1)/(2^(2))+….+(1)/(2^(n-1))`, find the least value of n for which `2- S_(n) lt (1)/(100)`.

Answer

To solve the problem, we start with the given series:

\[ S_n = 1 + \frac{1}{2} + \frac{1}{2^2} + \ldots + \frac{1}{2^{n-1}} \]

This is a geometric series where the first term \(a = 1\) and the common ratio \(r = \frac{1}{2}\).

### Step 1: Find the sum \(S_n\)

The formula for the sum of the first \(n\) terms of a geometric series is given by:

\[ S_n = \frac{a(1 - r^n)}{1 - r} \]

Substituting the values of \(a\) and \(r\):

\[ S_n = \frac{1(1 - (\frac{1}{2})^n)}{1 - \frac{1}{2}} = \frac{1 - \frac{1}{2^n}}{\frac{1}{2}} = 2(1 - \frac{1}{2^n}) = 2 - \frac{2}{2^n} \]

### Step 2: Set up the inequality

We need to find the least value of \(n\) such that:

\[ 2 - S_n < \frac{1}{100} \]

Substituting \(S_n\) into the inequality:

\[ 2 - \left(2 - \frac{2}{2^n}\right) < \frac{1}{100} \]

This simplifies to:

\[ \frac{2}{2^n} < \frac{1}{100} \]

### Step 3: Solve the inequality

To eliminate the fraction, we can multiply both sides by \(100 \cdot 2^n\):

\[ 200 < 2^n \]

### Step 4: Find the least integer \(n\)

Now, we need to find the smallest integer \(n\) such that \(2^n > 200\).

Calculating powers of \(2\):

- \(2^6 = 64\)
- \(2^7 = 128\)
- \(2^8 = 256\)

Since \(2^8 = 256\) is the first power of \(2\) greater than \(200\), we find that:

\[ n = 8 \]

### Conclusion

Thus, the least value of \(n\) for which \(2 - S_n < \frac{1}{100}\) is:

\[ \boxed{8} \]