Question

1 + 1⁄2 + 1⁄4 + 1⁄8 + 1⁄16 + 1⁄32 + 1⁄64. . . Notice that the denominator of each fraction in the sum is twice the denominator that comes before it. If you continue adding on fractions according to this pattern, when will you reach a sum of 2?

Answer 1

Multiplying by 2 is the rule...plz mark me as brainliest as just joined!

Answer 2

Answer:

$n=\infty$

Step-by-step explanation:

The given series is

$1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32} +\frac{1}{64} +...$

This is a geometric series and the first term is $a_1=1$

The common ratio is $r=\frac{a_2}{a_1}$

$\Rightarrow r=\frac{\frac{1}{2}}{1}=\frac{1}{2}$

The sum of the first n terms of this geometric series is given by,

$S_n=\frac{a_1(r^n-1)}{r-1}$

We want to find $n$ such $S_n=2$.

This implies that,

$2=\frac{1((\frac{1}{2})^n-1)}{\frac{1}{2}-1}$

We simplify to get,

$2=\frac{(\frac{1}{2})^n-1}{-\frac{1}{2}}$

We multiply through by $-\frac{1}{2}$ to get,

$-1=(\frac{1}{2})^n-1}$

We group the constant terms to get,

$-1+1=(\frac{1}{2})^n}$

This implies that,

$0=(\frac{1}{2})^n}$

We apply the laws of indices to further obtain,

$0=\frac{1}{2^n}}$

$\Rightarrow 2^n=\frac{1}{0}}$

We got division by zero, this means that $n$  is infinity.

$2$ is the sum to infinity of the given series.

Hence we do not have a finite number of terms.