Question

the sum of the first 20 terms of series 1+3/2+7/4+15/8+31/16+...... is

Answer 1

a= 1 ,, d= 3/2-1= -1/2 ,,,, find S20

Answer 2

Answer:

$\sum(1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+...)=38.0000019073$                

Step-by-step explanation:

Given : Series $1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+...$

To find : The sum of first 20 terms of series ?

Solution :

Series $1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+...$

We re-write the series as,

$=1+(2-\dfrac{1}{2^1})+(2-\dfrac{1}{2^{2}})+(2-\dfrac{1}{2^{3}})+....+(2-\dfrac{1}{2^{19}})$

$=1+2\times19-\dfrac{1}{2}(1+\dfrac{1}{2}+\dfrac{1}{2^{2}}+....+\dfrac{1}{2^{18} })$

Now, The bracket form a geometric finite series.

The formula of sum of finite geometric series is  

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

Here, $r=\frac{1}{2}$, $a=\frac{1}{2}$

Applying,

$=1+38-\dfrac{1}{2}(\dfrac{1-\dfrac{1}{2^{19}}}{1-\dfrac{1}{2}})$

$=39-\frac{1}{2}(\dfrac{2^{19}-1 }{2^{17}})$

$=38.0000019073$

Therefore, The sum of the first 20 terms of series is

$\sum(1+\frac{3}{2}+\frac{7}{4}+\frac{15}{8}+\frac{31}{16}+...)=38.0000019073$