Question

How is the sum expressed in sigma notion? Multiple choice

Answer 1

Answer:

The given sum expressed in sigma notation is

$\sum\limits_{i=1}^{5}4^{i-4}$

Therefore $\sum\limits_{i=1}^{5}4^{i-4}=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4$

Step-by-step explanation:

Given series is $\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4$

The given sum expressed in sigma notation is

$\sum\limits_{i=1}^{5}4^{i-4}$

• Now verify that the sigma notation $\sum\limits_{i=1}^{5}4^{i-4}$ is correct or not
• Now expand the series
• $\sum\limits_{i=1}^{5}4^{i-4}=4^{1-4}+4^{2-4}+4^{3-4}+4^{4-4}+4^{5-4}$  
• $=4^{-3}+4^{-2}+4^{-1}+4^{0}+4^{1}$  ( using the properties $a^{-m}=\frac{1}{a^m}$ and $a^0=1$ )

• $=\frac{1}{4^3}+\frac{1}{4^2}+\frac{1}{4^1}+1+4$
• $=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4$

Therefore

$\sum\limits_{i=1}^{5}4^{i-4}=\frac{1}{64}+\frac{1}{16}+\frac{1}{4}+1+4$

• Hence verified

Answer 2

Answer:

Option 2

Step-by-step explanation:

1/64 = 4^-3

1/16 = 4^-2

1/4 = 4^-1

1 = 4^0

4 = 4^1