Question

Find the sum: $\frac{1}{2}+\frac{1}{4}+.............+\frac{1}{2^{10}}$

Answer 1

It is given that the first term of G.P is $a_1=\frac{1}{2}$ and the second term is $a_2=\frac{1}{4}$, therefore,

$r=\frac{\frac{1}{4}}{\frac{1}{2}}=\frac{1}{4}\times 2=\frac{1}{2}$

We know that the formula for the $n$th term of an G.P is $T_n=a{r}^{n-1}$, where $a$ is the first term, $r$ is the common ratio.

It is given that the first term of G.P is $a=\frac{1}{2}$, the $n$th term is $T_n=\frac{1}{2^{10}}$ and the common ratio is $r=\frac{1}{2}$, therefore,

$T_n=a{r}^{n-1}\Rightarrow \frac{1}{2^{10}}=\left(\frac{1}{2}\right)\times {\left(\frac{1}{2}\right)}^{n-1}\Rightarrow \frac{1}{2^{10}}={\left(\frac{1}{2}\right)}^{n-1+1}\Rightarrow \frac{1}{2^{10}}={\left(\frac{1}{2}\right)}^n\Rightarrow \frac{1}{2^{10}}=\frac{1}{2^n}\Rightarrow n=10$

Now, the formula for sum of $n$ terms of G.P is $S_n=\frac{a(1-r^n)}{1-r}$, where $a$ is the first term, $r$ is the common ratio.

In the given geometric series, the first term is $a=\frac{1}{2}$, the common ratio is $r=\frac{1}{2}$ and the number of terms are $n=10$, therefore,

$S_{10}=\frac{\left(\frac{1}{2}\right)(1-{\left(\frac{1}{2}\right)}^{10})}{1-\frac{1}{2}}=\frac{\left(\frac{1}{2}\right)(1-\frac{1}{2^{10}})}{\frac{1}{2}}=1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{10}}=\frac{1024-1}{1024}=\frac{1023}{1024}$

Hence, the sum is $\frac{1023}{1024}$.