Question

How is the sum expressed in sigma notion? Multiple choice

Answer 1

Option 2

Explanation:

1/64 = 4^-3

1/16 = 4^-2

1/4 = 4^-1

1 = 4^0

4 = 4^1

Answer 2

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)=(1)/(64)+(1)/(16)+(1)/(4)+1+4`

Explanation:

Given series is
`(1)/(64)+(1)/(16)+(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)=(1)/(a^m)` and
  `a^0=1` )

• 
  `=(1)/(4^3)+(1)/(4^2)+(1)/(4^1)+1+4`
• 
  `=(1)/(64)+(1)/(16)+(1)/(4)+1+4`

Therefore

`\sum\limits_(i=1)^(5)4^(i-4)=(1)/(64)+(1)/(16)+(1)/(4)+1+4`

• Hence verified