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 Post subject: Great practice questionPosted: Thu Jul 29, 2010 4:42 pm

Joined: Thu Jul 03, 2008 2:13 pm
Posts: 117
Hello everyone:

A student of mine just emailed me with a question that I think could help everyone, so let me post my response:

What is: 2+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8 = ??

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My response:

Wow – great question!

If we start with the statement as given:

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

We’re adding a bunch of exponential terms together, which we’re not good at. So let’s try to combine a few of them to see if we can start factoring and multiplying (which we do pretty well):

2 + 2 = 4, so we can replace those two terms with 4, or 2^2

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

2^2 + 2^2 is the same as 2(2^2), and we can express that as 2^3, so we can replace those two terms:

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

Now we can do the same with 2^3 + 2^3 – it’s 2(2^3), which is 2^4, which lets us replace that and leaves us with:

2^4+2^4+2^5+2^6+2^7+2^8

By now you may have seen the pattern – we can replace the first two terms with the third term each time, because the first two terms add up to 2*one of them, and that gets us to the same as the third, which replicates the cycle. If we keep doing that:

2^4 + 2^4 = 2(2^4) = 2^5, leaving

2^5 + 2^5 + 2^6 + 2^7 + 2^8

We can replace those first two 2^5 terms with 2^6 leaving:

2^6 + 2^6 + 2^7 + 2^8

Again, replace the first two with the third:

2^7 + 2^7 + 2^8

Do that one more time to get:

2^8 + 2^8 and that leaves us with:

2(2^8) = 2^9

The answer is 2^9

Once you notice the pattern you can work through this one that much faster, so remember that when dealing with exponents you should:

• Try to multiply whenever possible (factor!) and not add
• Look for patterns

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 Post subject: Re: Great practice questionPosted: Thu Aug 05, 2010 1:08 pm

Joined: Wed Aug 04, 2010 6:03 am
Posts: 2
That is a great explanation and it reminds me of a concept I studied at school...

2^0 + 2^1 + 2^2 + .... + 2^n = 2^(n + 1) - 1

All the consecutive powers of 2 add up to give a result that is 1 less than its next power.. e.g.
2^0 + 2^1 + 2^2 +....+ 2^8 = 2^9 - 1

In fact, we can add up different powers of 2 to get all numbers from 1 to 2^ (n - 1) - 1...
Lets say we have 2^0, 2^1, 2^2 and 2^3. We can add up these four numbers (1, 2, 4, 8) in different ways to get all integers from 1 to 15.
1 + 2 = 3
1 + 4 = 5
2 + 4 = 6
1 + 2 + 4 = 7
1 + 8 = 9
2 + 8 = 10
1 + 2 + 8 = 11
4 + 8 = 12
1 + 4 + 8 = 13
2 + 4 + 8 = 14
1 + 2 + 4 + 8 = 15
To get a 16, you will now need to introduce 2^4.

Such a relation holds for other numbers too...

2( 3^0 + 3^1 + 3^2 + ....+ 3^n) = 3^(n + 1) - 1
3( 4^0 + 4^1 + 4^2 + ....+ 4^n) = 4^(n + 1) - 1
and so on...

So now, a question based on this: A small fruit vendor wants to sell fruit in multiples of 1 pound. He has a traditional weighing balance which has two pans, one for the fruit to be weighed and other for the metal weights. What is the minimum number of weights he needs to have to be able to sell upto 100 pounds of fruit in a single weighing?

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