PLEASE HELP!!! i’ve been stuck on this for so long!!!

Determine the number of terms in the geometric series 2+8+32+...+2048

Answer:

6 terms

Step-by-step explanation:

we have

[tex]2+8+32+..+2048[/tex]

we need to identify a pattern so

from 2 to 8 what could have happened?

well you could have added 6 and get 8

so that means that to 8 you have to add 6 and get 32

WRONG

so adding is not an option

what else could have happened?

well you could have multiplied by 4 and get 8

so that means that to 8 you have to multiply by 4 and get 32

which is true

so the geometric series is

[tex]2+4(2)+4(4(2))+4(4(4(2)))+...+[/tex] what?

you can see that in the 2nd term we have 1 four, in the 3rd we have 2 fours

so we can conclude in the nth term we have n-1 fours

so the formula is

[tex]2+8+32+...+4^{n-1} *2[/tex]

so we need to determine when

[tex]4^{n-1}*2=2048[/tex]

[tex]4^{n-1}*2*\frac{1}{2} =2048*\frac{1}{2} \\\\4^{n-1}=1024[/tex]

if we factorize 4 as [tex]2^2[/tex] and 1024 as [tex]2^{10}[/tex]

we have

[tex](2^2)^{n-1}=2^{10}\\\\2^{2(n-1)}=2^{10}\\\\2^{2n-2}=2^{10}[/tex]

so we see that the bases are equal, that must mean the exponents are equal

[tex]2n-2=10\\\\2n=10+2\\\\2n=12\\\\n=6[/tex]

so there are 6 terms

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