Question

(Picture) INFINITE SEQUENCES AND SERIES PLEASE HELP!!

Answer 1

Option d. 32/5

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

Answer 2

32/5

option D

Explanation:

Given series is 8 -2 + 1/2 - 1/8+.....

Given series is geometric

to find sum we use formula

`s_n=(a)/(1-r)`

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

`(-2)/(8) =(-1)/(4)`

Now plug in r=-1/4 and a= 8 in the formula to find sum

`s_n=(a)/(1-r)`

`s_n=(8)/(1-((-1)/(4)))`

`s_n=(8)/(1+((1)/(4)))`

`s_n=(8)/(((5)/(4)))`

Flip the bottom fraction and multiply with 8

`s_n=8*((4)/(5))`

sum = 32/5