Question

Find the sum of given series 1 + 2 + 4 + 8 +..............+ 256

Answer 1

Explanation:

a = 1 , r = 2

An = ar^n-1

256 = (1)*(2)^n-1

(2)^8 = (2)^n-1

n - 1 = 8

n = 9

Sum of this Series

S9 = 1(2^9 - 1)/(2 - 1)

= 1(2^9 - 1) / 1

= 1 (512 - 1)

= 1 × 511

= 511

I hope it will help you

Answer 2

Given:

A series is provided as follows :

$1 + 2 + 4 + 8 \: . \: . \: . \: . + 256$

To find:

Sum of the series ?

Calculation:

• We can easily understand that the series is a GEOMETRIC PROGRESSION with 1st term as 1 and common ratio as 2 .

First, let's find out the number of terms in the series :

$\therefore \: T = a {r}^{n - 1}$

$\implies \: 256 = 1 \times {2}^{n - 1}$

$\implies \: 256 = {2}^{n - 1}$

$\implies \: {2}^{8} = {2}^{n - 1}$

$\implies \: n - 1 = 8$

$\implies \: n = 9$

Now, sum of series is :

$S = \dfrac{a( {r}^{n} - 1) }{r - 1}$

$\implies \: S = \dfrac{1( {2}^{9} - 1) }{2 - 1}$

$\implies \: S = \dfrac{1(512 - 1)}{1}$

$\implies \: S = 511$

So, sum of series is 511.