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Dima020 [189]
3 years ago
6

What is the sum of the geometric series 2^0 + 2^1 + 2^2 + 2^3 + 2^3 + 2^4 + … + 2^9?

Mathematics
2 answers:
avanturin [10]3 years ago
5 0
So this one you don't multiply the base by the exponent. in this case its 2 times two how ever many time the exponent says.
2^0= 2
2^1= 2
2^2=4
2^3= 8
2^4=16
2^5=32
2^6=64
2^7=128
2^8=256
2^9=512
 So then you add all them up 
2+2+4+8+16+32+64+128+256+512= 1024

so there is your answer 1024


GREYUIT [131]3 years ago
4 0
Sum is
S_{n}=\frac{a_{1}(1-r^{n})}{1-r}

r=common ratio
a1=first term
it looks like 2^0=1 is the first term aka a1
it goes to the 9th term (2^9)

sub
S_{9}=\frac{1(1-(2)^{9})}{1-2}
S_{9}=\frac{1-512}{-1}
S_{9}=\frac{-511}{-1}
S_{9}=511

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lakkis [162]

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  • y = 4

Step-by-step explanation:

<u>We know that:</u>

  • 4x + y = 14
  • 6x - 3y = 3

<u>Let's first solve for y by choosing any equation.</u>

  • 4x + y = 14
  • => y = 14 - 4x

<u>Now, let's substitute the value of y into the second equation.</u>

  • 6x - 3y = 3
  • => 6x - 3(14 - 4x) = 3
  • => 6x - 42 + 12x = 3
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<u>Now, let's substitute the value of x into the equation of y.</u>

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