(Picture) INFINITE SEQUENCES AND SERIES PLEASE HELP!!
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Answer:
32/5
option D
Step-by-step explanation:
Given series is 8 -2 + 1/2 - 1/8+.....
Given series is geometric
to find sum we use formula
[tex]s_n=\frac{a}{1-r}[/tex]
where 'a' is the first term
and r is the common ratio
to find out common ratio 'r' we divide second term by first term
[tex]\frac{-2}{8} =\frac{-1}{4}[/tex]
Now plug in r=-1/4 and a= 8 in the formula to find sum
[tex]s_n=\frac{a}{1-r}[/tex]
[tex]s_n=\frac{8}{1-(\frac{-1}{4})}[/tex]
[tex]s_n=\frac{8}{1+(\frac{1}{4})}[/tex]
[tex]s_n=\frac{8}{(\frac{5}{4})}[/tex]
Flip the bottom fraction and multiply with 8
[tex]s_n=8*(\frac{4}{5})[/tex]
sum = 32/5
Answer:
Option d. 32/5
Step-by-step explanation:
a0=a=8
a1=-2
a2=1/2
a3=-1/8
a1/a0=(-2)/8=-2/8=-(2/2)/(8/2)→a2/a1=-1/4
a2/a1=(1/2)/(-2)=(1/2)*(-1/2)=-(1*1)/(2*2)→a3/a2=-1/4
a3/a2=(-1/8)/(1/2)=(-1/8)*(2/1)=-(1*2)/(8*1)=-2/8=-(2/2)/(8/2)→a4/a3=-1/4
r=a1/a0=a2/a1=a3/a2→r=-1/4
Absolute value of r: !r!=!-1/4!=1/4=0.25<1, then the series converges
Sum of the geometric series: S=?
S=a/(1-r)
Replacing the known values:
S=8/[1-(-1/4)]
S=8/(1+1/4)
S=8/[(1/1)*(4/4)+1/4]
S=8/[(1*4)/(1*4)+1/4]
S=8/(4/4+1/4)
S=8/[(4+1)/4]
S=8/(5/4)
S=(8/1)*(4/5)
S=(8*4)/(1*5)
S=32/5