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Please help meeeeeeeeeee

Please help meeeeeeeeeee-example-1

1 Answer

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Answer:


(468)/(25)

Explanation:

We are asked to find


\displaystyle \sum _(n=0)^33\left((1)/(5)\right)^(n-1)

We are asked to find s₄ which is the sum of the first 4 terms of the series. Since n starts at 0, this would be
a₀ + a₁ + a₂ + a₃

We can compute the individual terms and add them up

n = 0 \\\rightarrow \displaystyle a_0 = 3\left((1)/(5)\right)^(0-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(-1)\\\\\rightarrow \displaystyle 3\left(5}\right)^(1)\\ \rightarrow \displaystyle 15

This would be the first term of the geometric sequence

n = 1

\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(1-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(0)\\\\\rightarrow \displaystyle 3\cdot 1}\\ \rightarrow \displaystyle a_1 = 3

n = 2

\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(2-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(1)\\\\\rightarrow \displaystyle a_2 = (3)/(5)

n = 3

\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(3-1)\\\rightarrow \displaystyle 3\left((1)/(5)\right)^(2)\\\\\rightarrow \displaystyle a_3 = (3)/(25)

Adding these we get

15 + 3 + 3/5 + 3/25 =
(468)/(25)

Alternate route
In a geometric sequence the sum of the formula for the first n terms will be


a_0 + a_1(1-r^(n))/(1-r)

We computed a₀ as 15 and a₁ as 3

So sum


= 15 + 3\left((1-\left((1)/(5)\right)^3)/(1-(1)/(5))\right)\\

If you evaluate this, it should work out to the same fraction:

(468)/(25)

User Janusz Lenar
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