What is the sum of the geometric sequence 1, 4, 16, … if there are 8 terms?

Question

What is the sum of the geometric sequence 1, 4, 16, … if there are 8 terms?

1 Answer

Impact of this question

10039 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-01-25T03:45:36

$\frac{a_{2}}{a_{1}} = \frac{4}{1} = 4$ $a_2/a_1=4/1=4$

$\frac{a_{3}}{a_{2}} = \frac{16}{4} = 4$ $a_3/a_2=16/4=4$

$\Rightarrow$ $implies$ common ratio $= r = 4$ $=r=4$ and $a_{1} = 1$ $a_1=1$

Sum of geometric series is given by

$S u m = \frac{a_{1} (1 - r^{n})}{1 - r}$ $Sum=(a_1(1-r^n))/(1-r)$

Where $n$ $n$ is number of terms, $a_{1}$ $a_1$ is the first term, $r$ $r$ is the common ratio.

Here $a_{1} = 1$ $a_1=1$ , $n = 8$ $n=8$ and $r = 4$ $r=4$

$\Rightarrow S u m = \frac{1 (1 - 4^{8})}{1 - 4} = \frac{1 - 65536}{- 3} = \frac{- 65535}{- 3} = 21845$ $implies Sum=(1(1-4^8))/(1-4)=(1-65536)/-3=(-65535)/-3=21845$

Hence, the sum is $21845$ $21845$

Science

Math

Humanities

... and beyond

What is the sum of the geometric sequence 1, 4, 16, … if there are 8 terms?

1 Answer

Related questions

Impact of this question