Question

Evaluate the summation of 3 times negative 2 to the n minus 1 power, from n equals 1 to 5..

Answer 1

ANSWER

$\sum_{n=1}^53(-2)^{n-1} = 33$

EXPLANATION

The given series is

$\sum_{n=1}^53(-2)^{n-1}$

This is a geometric series with general term,

$a_n=3( { - 2)}^{n - 1}$

The terms are:

$a_1=3( { - 2)}^{1- 1} = 3$

$a_2=3( { - 2)}^{2- 1} = - 6$

$a_3=3( { - 2)}^{3- 1} = 12$

$a_4=3( { - 2)}^{4 - 1} = - 24$

$a_5=3( { - 2)}^{5- 1} = 48$

This implies that,

$\sum_{n=1}^53(-2)^{n-1} = 3 + - 6 + 12 + - 24 + 48$

$\sum_{n=1}^53(-2)^{n-1} = 33$