Answer:
S∞ = 7 Option C
Step-by-step explanation:
Given that:
∑21[tex](\frac{1}{4} )^{i}[/tex] i = 1 to ∞
We need to find the sum.
Firstly we will find the sequence
a₁ = [tex]21(\frac{1}{4} )^{1} = (\frac{21}{4} )[/tex]
a₂ = [tex]21(\frac{1}{4} )^{2} = (\frac{21}{16} )[/tex]
a₃ = [tex]21(\frac{1}{4} )^{3} = (\frac{21}{64} )[/tex]
a₄ = [tex]21(\frac{1}{4} )^{4} = (\frac{21}{256} )[/tex]
this sequence is geometric
So, we use sum of infinite terms
a = [tex]\frac{21}{4}[/tex]
r = [tex]\frac{\frac{21}{16} }{\frac{21}{4} } = \frac{1}{4}[/tex]
The sum of infinite terms of a GP series
S∞ = [tex]\frac{a}{1-r}[/tex]
where a is first term
r is common ratio and 0<r<1
S∞ = [tex]\frac{\frac{21}{4} }{1 - \frac{1}{4} } = \frac{\frac{21}{4} }{\frac{3}{4} }[/tex]
S∞ = 7
