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elena-14-01-66 [18.8K]

3 years ago

6

Sum of series 1 +1/3 +1/3².... is?

1 answer:

Charra [1.4K]3 years ago

7 0

Answer:

The sum of the series is 3/2

Step-by-step explanation:

Given

1 + 1/3 + 1/3^2 + ....

Required

The sum of the series

This implies that we calculate the sum to infinity.

We have:

$a = 1$ -- The first term

First, calculate the common ratio (r)

$r = \frac{1}{3^2} / \frac{1}{3}$

Change to product

$r = \frac{1}{3^2} * \frac{3}{1}$

Solve

$r = \frac{1}{3}$

The sum of the series is then calculated as:

$S_{\infty} = \frac{a}{1 - r}$

$S_{\infty} = \frac{1}{1 - 1/3}$

Solve the denominator

$S_{\infty} = \frac{1}{2/3}$

Express as product

$S_{\infty} = 1 * \frac{3}{2}$

$S_{\infty} = \frac{3}{2}$

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An x-method chart shows the product a c at the top of x and b at the bottom of x. Above the chart is the expression a x squared

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Answer:

The other factor is (x + 2).

Step-by-step explanation:

If x - 10 is a factor of x^2 - 8x - 20 then we must find another factor that includes +2x, so that we have -10x + 2x = -8x. That factor is (x + 2):

x^2 - 8x - 20 = (x - 10)(x + 2)

As a test, multiply -10 and +2. We get -20, matching the constant of the given polynomial. Also, -10x + 2x = -8x, another match.

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3 years ago

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To finance her community college education, Sarah takes out a loan for $4300. After a year Sarah decides to pay off the inter

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Step-by-step explanation:

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3 years ago

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A^2 + b^2 + c^2 = 2(a − b − c) − 3. (1) Calculate the value of 2a − 3b + 4c.

Verdich [7]

Answer:

$2a - 3b + 4c = 1$

Step-by-step explanation:

Given

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Required

Determine $2a - 3b + 4c$

$a^2 + b^2 + c^2 = 2(a - b - c) - 3$

Open bracket

$a^2 + b^2 + c^2 = 2a - 2b - 2c - 3$

Equate the equation to 0

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 3 = 0$

Express 3 as 1 + 1 + 1

$a^2 + b^2 + c^2 - 2a + 2b + 2c + 1 + 1 + 1 = 0$

Collect like terms

$a^2 - 2a + 1 + b^2 + 2b + 1 + c^2 + 2c + 1 = 0$

Group each terms

$(a^2 - 2a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

Factorize (starting with the first bracket)

$(a^2 - a -a + 1) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$(a(a - 1) -1(a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1) (a - 1)) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + 2b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b^2 + b+b + 1) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + (b(b + 1)+1(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)(b + 1)) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + 2c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c^2 + c+c + 1) = 0$

$((a - 1)^2) + ((b + 1)^2) + (c(c + 1)+1(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)(c + 1)) = 0$

$((a - 1)^2) + ((b + 1)^2) + ((c + 1)^2) = 0$

Express 0 as 0 + 0 + 0

$(a - 1)^2 + (b + 1)^2 + (c + 1)^2 = 0 + 0+ 0$

By comparison

$(a - 1)^2 = 0$

$(b + 1)^2 = 0$

$(c + 1)^2 = 0$

Solving for $(a - 1)^2 = 0$

Take square root of both sides

$a - 1 = 0$

Add 1 to both sides

$a - 1 + 1 = 0 + 1$

$a = 1$

Solving for $(b + 1)^2 = 0$

Take square root of both sides

$b + 1 = 0$

Subtract 1 from both sides

$b + 1 - 1 = 0 - 1$

$b = -1$

Solving for $(c + 1)^2 = 0$

Take square root of both sides

$c + 1 = 0$

Subtract 1 from both sides

$c + 1 - 1 = 0 - 1$

$c = -1$

Substitute the values of a, b and c in $2a - 3b + 4c$

$2a - 3b + 4c = 2(1) - 3(-1) + 4(-1)$

$2a - 3b + 4c = 2 +3 -4$

$2a - 3b + 4c = 1$

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3 years ago

Write 0.000000364 in standard form

Thepotemich [5.8K]

Answer:

yes

Step-by-step explanation:

this number is a decimal

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2 years ago

If a+2b/a-b =2/3 then b/a=?

Dominik [7]

Answer:

$\frac{b}{a}$ = - $\frac{1}{8}$

Step-by-step explanation:

Given

$\frac{a+2b}{a-b}$ = $\frac{2}{3}$ ( cross- multiply )

2(a - b) = 3(a + 2b) ← distribute parenthesis on both sides )

2a - 2b = 3a + 6b ( subtract 3a from both sides )

- a - 2b = 6b ( add 2b to both sides )

- a = 8b ( divide both sides by 8 )

- $\frac{1}{8}$ a = b ( divide both sides by a )

- $\frac{1}{8}$ = $\frac{b}{a}$

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2 years ago

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