To find the sum to infinity of a geometric progression (G.P.), we use the formula for the sum of an infinite geometric series. The formula is given by:
[tex]\[ S_\infty = \frac{a}{1 - r} \][/tex]
where:
- [tex]\( a \)[/tex] is the first term of the series,
- [tex]\( r \)[/tex] is the common ratio of the series.
Let's solve each part step-by-step.
### Part (a)
For the series [tex]\( 1 + \frac{1}{4} + \frac{1}{16} + \ldots \)[/tex]:
1. Identify the first term [tex]\( a \)[/tex]:
[tex]\[ a = 1 \][/tex]
2. Determine the common ratio [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\frac{1}{4}}{1} = \frac{1}{4} \][/tex]
3. Substitute these values into the sum to infinity formula:
[tex]\[ S_\infty = \frac{1}{1 - \frac{1}{4}} \][/tex]
4. Simplify the expression:
[tex]\[ S_\infty = \frac{1}{\frac{3}{4}} = \frac{1 \times 4}{3} = \frac{4}{3} = 1.3333333333333333 \][/tex]
So, the sum to infinity for part (a) is:
[tex]\[ S_\infty = 1.3333333333333333 \][/tex]
### Part (b)
For the series [tex]\( 1 + \frac{9}{10} + \frac{81}{100} + \ldots \)[/tex]:
1. Identify the first term [tex]\( a \)[/tex]:
[tex]\[ a = 1 \][/tex]
2. Determine the common ratio [tex]\( r \)[/tex]:
[tex]\[ r = \frac{\frac{9}{10}}{1} = \frac{9}{10} \][/tex]
3. Substitute these values into the sum to infinity formula:
[tex]\[ S_\infty = \frac{1}{1 - \frac{9}{10}} \][/tex]
4. Simplify the expression:
[tex]\[ S_\infty = \frac{1}{\frac{1}{10}} = \frac{1 \times 10}{1} = 10 \][/tex]
So, the sum to infinity for part (b) is:
[tex]\[ S_\infty = 10.000000000000002 \][/tex]
Therefore, the sum to infinity for the given series are:
- Part (a): [tex]\( 1.3333333333333333 \)[/tex]
- Part (b): [tex]\( 10.000000000000002 \)[/tex]
