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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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Middle School Math Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

Grades 5-8, Ages 10-13, MATHCOUNTS, AMC 8

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#1 h Jul 10, 2016, 1:35 PM • 2 Y

Y by Adventure10, Mango247

What is the sum of the infinite series 1 + 2/3 + 4/9 + 8/27. . .?

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#2 h Jul 10, 2016, 1:35 PM • 4 Y

Y by Mathisfun04, blitzkrieg21, Adventure10, Mango247

$\boxed{3}$ , how is this a MC Nats prob?

This post has been edited 1 time. Last edited by claserken, Jul 10, 2016, 1:36 PM

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First

#3 h Jul 10, 2016, 1:35 PM • 4 Y

Y by Mathisfun04, blitzkrieg21, Adventure10, Mango247

3, this problem is fairly trivial
Sniped

This post has been edited 1 time. Last edited by First, Jul 10, 2016, 1:36 PM

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#4 h Jul 10, 2016, 2:06 PM • 2 Y

Y by Adventure10, Mango247

claserken wrote:

$\boxed{3}$ , how is this a MC Nats prob?

This is a bit harder than most MC nats sprint 1s, don't you agree?

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#5 h Jul 10, 2016, 2:08 PM • 2 Y

Y by Adventure10, Mango247

mathguy5041 wrote:

claserken wrote:

$\boxed{3}$ , how is this a MC Nats prob?

This is a bit harder than most MC nats sprint 1s, don't you agree?

I've never taken a MC Nat test before.

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#6 h Jul 10, 2016, 3:25 PM • 1 Y

Y by Adventure10

claserken wrote:

mathguy5041 wrote:

claserken wrote:

$\boxed{3}$ , how is this a MC Nats prob?

This is a bit harder than most MC nats sprint 1s, don't you agree?

I've never taken a MC Nat test before.

MC Nat Sprint #1 are supposed to be of that material, a bit easy perhaps.

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asp211

#7 h Jul 10, 2016, 5:35 PM • 2 Y

Y by Adventure10, Mango247

it could be countdown

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hinna

#8 h Jul 10, 2016, 7:29 PM • 4 Y

Y by The_Herring, Blue_Whale, Adventure10, Mango247

Well its 3. I guess I'll explain why:

This is a geometric series with common ratio $\frac {2}{3}$ and first term 1. To find the sum of an infinite geometric series with common term $r$ and 1st term $n$ , we do $\frac {n}{1-r}$ and plugging the numbers in, we get $\frac {1}{1-\frac {2}{3}}=\boxed 3$ .

The proof for the sum formula is that if you have $a_1, a_2, a_3...$ where each term is multiplied by a common factor. Then we write $a_1, ra_1, r^2a_1...=S$ . Then $Sr=ra_1+r^2a_1...$ which means the sum times the common factor between two consecutive terms plus the first term equals the sum is : $S=a_1+Sr$ . AKA simplifying, we get:
$S-Sr=a_1$
$S\cdot(1-r)=a_1$
$S=\frac {a_1}{1-r}$

This post has been edited 1 time. Last edited by hinna, Jul 10, 2016, 7:41 PM
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#9 h Aug 19, 2016, 9:30 PM • 2 Y

Y by Adventure10, Mango247

Mathisfun04 wrote:

What is the sum of the infinite series 1 + 2/3 + 4/9 + 8/27. . .?

This post has been edited 1 time. Last edited by blitzkrieg21, Aug 19, 2016, 9:31 PM

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#10 h Aug 19, 2016, 9:44 PM • 2 Y

Y by Adventure10, Mango247

@claserken

If you've never taken a MC Nats test before, then how could you possibly ask, "how is this a MC Nats prob?"

Also, using your logic, if I've never taken an AMC 12 test before, and I know that AMC 12 is harder than AMC 10, am I right to assume that there will be no simple questions on the AMC 12 at all?

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jack74

#11 h Aug 19, 2016, 9:47 PM • 1 Y

Y by Adventure10

There is no way that this is the sum. This may be the partial sum, but definitely not the real sum.

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#12 h Aug 19, 2016, 10:13 PM • 2 Y

Y by Adventure10, Mango247

@jack74
If that's the case, then what is the "real sum"?

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#13 h Aug 19, 2016, 10:34 PM • 2 Y

Y by Adventure10, Mango247

jack74 wrote:

There is no way that this is the sum. This may be the partial sum, but definitely not the real sum.

Your profile says that you are in elementary school, so I'll assume so. Maybe you have not learned this yet, but it is possible to sum up an infinite number of elements in a series with a common ratio less than 1. We say that the series converges to that number.

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#14 h Aug 20, 2016, 1:14 AM • 2 Y

Y by Adventure10, Mango247

jack74 wrote:

There is no way that this is the sum. This may be the partial sum, but definitely not the real sum.

You want proof? I'll give you proof. (Because everyone else hasn't hid their solutions, it's no use hiding mine.)

Note that we can assign the value $s$ to $1 + \dfrac23 + \dfrac49 + \cdots.$ We can factor out a term of $\dfrac23$ out of every term after $1:$

$s = 1 + \dfrac23\left(1 + \dfrac23 +\dfrac49\right).$
The expression in the parentheses is just $s$ ! For a finite series, this would be untrue. However, this is an infinite series. We can do things like this to them.

Now, just substitute:

$s = 1+\dfrac23(s).$
And solve:

$\dfrac13(s) = 1.$
Thus $\boxed{s=3.}$

If you still don't believe me, I'll give you a piece of very wise advice. Never, ever take calculus.

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#15 h Aug 20, 2016, 3:39 AM • 2 Y

Y by Adventure10, Mango247

jack74 wrote:

There is no way that this is the sum. This may be the partial sum, but definitely not the real sum.

What is real sum?

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#16 h Aug 20, 2016, 3:56 AM • 2 Y

Y by Adventure10, Mango247

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#19 h Aug 20, 2016, 5:14 AM • 2 Y

Y by Adventure10, Mango247

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#20 h Aug 20, 2016, 4:35 PM • 3 Y

Y by dragonmaster3000, Adventure10, Mango247

Mathisfun04 wrote:

What is the sum of the infinite series 1 + 2/3 + 4/9 + 8/27. . .?

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