I have an equation as below:

3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9 + 3^10

Note: ^ stands for exponent

Anybody knows the shortcut to calculate that equation i/o multiplying and plus one by one. Please advise

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I have an equation as below:

3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9 + 3^10

Note: ^ stands for exponent

Anybody knows the shortcut to calculate that equation i/o multiplying and plus one by one. Please advise

3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9 + 3^10

Note: ^ stands for exponent

Anybody knows the shortcut to calculate that equation i/o multiplying and plus one by one. Please advise

- ice_breaker
**Posts:**6**Joined:**Sat Jan 16, 2010 3:40 am**Reputation:****0**

Thank you Teacher.

I did have a look at your given site and some sites thru Google. Finally, I figure out the solution as below:

3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9 + 3^10

1) Get out a common factor

3(1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^9)

2) Use the formula [(a^n - b^n)/2] , as a=3, b=1, to calculate the parenthesis, then multiply by 3

3 * [(3^10 - 1^10)/2] = 88 572.

The result is 88,572.

Is this correct Teacher? Please advise

I did have a look at your given site and some sites thru Google. Finally, I figure out the solution as below:

3 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^8 + 3^9 + 3^10

1) Get out a common factor

3(1 + 3^2 + 3^3 + 3^4 + 3^5 + 3^6 + 3^7 + 3^9)

2) Use the formula [(a^n - b^n)/2] , as a=3, b=1, to calculate the parenthesis, then multiply by 3

3 * [(3^10 - 1^10)/2] = 88 572.

The result is 88,572.

Is this correct Teacher? Please advise

- ice_breaker
**Posts:**6**Joined:**Sat Jan 16, 2010 3:40 am**Reputation:****0**

I do not think that here we can use that formula.

Where you found it? Could I see, please?

The geometric series formula is:

[tex]S = a1 \frac{(1-q^n)}{1-q}[/tex]

a1 = 3 - the first number of the sequence

q = 3 - common ratio (because every number is the previous number multiplied by 3)

n=10

so

the sum is

[tex]Sum = 3 \frac{(1-3^{10})}{1-3} = 3 \frac{(1-3^{10})}{1-3}[/tex]

[tex]Sum = 3 \frac{(-59048)}{-2} = 88572[/tex]

Where you found it? Could I see, please?

The geometric series formula is:

[tex]S = a1 \frac{(1-q^n)}{1-q}[/tex]

a1 = 3 - the first number of the sequence

q = 3 - common ratio (because every number is the previous number multiplied by 3)

n=10

so

the sum is

[tex]Sum = 3 \frac{(1-3^{10})}{1-3} = 3 \frac{(1-3^{10})}{1-3}[/tex]

[tex]Sum = 3 \frac{(-59048)}{-2} = 88572[/tex]

- Math Tutor
- Site Admin
**Posts:**410**Joined:**Sun Oct 09, 2005 11:37 am**Reputation:****28**

This formula is provided by a friend of mine. He told me to use that formula to solve this equation.

I think I'd better use yours to standardize my knowledge. By the way, if the equation is like:

1/2 + 1/2^2 + 1/2^3 + 1/2^4 + ..... + 1/2^10

Can I use your given formula? Pls advise

I think I'd better use yours to standardize my knowledge. By the way, if the equation is like:

1/2 + 1/2^2 + 1/2^3 + 1/2^4 + ..... + 1/2^10

Can I use your given formula? Pls advise

- ice_breaker
**Posts:**6**Joined:**Sat Jan 16, 2010 3:40 am**Reputation:****0**

I do not think that here we can use that formula.

Where you found it? Could I see, please

Where you found it? Could I see, please

Hey,

it was very helpful, though I have an exam tomorrow, and the only thing I'm confused is the steps multiplying fractions like:

Image

It would be awesome if someone would show me steps (for beginners), I do know how to cross-multiply, it's just the n's confuse me.

it was very helpful, though I have an exam tomorrow, and the only thing I'm confused is the steps multiplying fractions like:

Image

It would be awesome if someone would show me steps (for beginners), I do know how to cross-multiply, it's just the n's confuse me.

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