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Find the sum of the first 8 terms of the following series, to the nearest integer.

144, 36,9, ...

2 Answers

10 votes

answer:

192

(the pattern is dividing each number by 4, if you need it)

User Pavel Tupitsyn
by
7.9k points
9 votes

Sure, I'd be happy to walk you through how to solve this problem.

This particular sequence of numbers is known as a geometric series. A geometric series is a series of numbers where each term after the first is found by multiplying the previous term by a fixed, non-zero number called the common ratio. In this case, it is easy to see that the common ratio is 1/4 because each term is a quarter of the previous one.

The first term (denoted by 'a') is 144, the number of terms (denoted by 'n') is 8, and the common ratio (denoted by 'r') is 1/4.

The sum ('S') of the first n terms of a geometric series can be found using the formula:

S_n = a * (1 - r^n) / (1 - r)

Let's plug in the values.

S_8 = 144 * (1 - (1/4)^8) / (1 - 1/4)

Perform the exponent first:

(1/4)^8 = 0.0000152587890625

Now substitute this back into the equation:

S_8 = 144 * (1 - 0.0000152587890625) / (1 - 1/4)

Subtract within the parentheses:

144 * (0.9999847412109375) / (0.75)

Multiply within the parentheses on top:

S_8 = 143.99704229736328 / 0.75

Now divide:

S_8 = 191.99605639648438

We round this to the nearest integer because the problem asks for the nearest integer.

S_8 = 192

So, the sum of the first 8 terms of the given series, rounded to the nearest integer, is .

Answer:192

User Atanisha
by
7.7k points

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