Before this post, I have done posts involving multiplication and squaring tricks. But today, I am going to talk about some addition.

There are ways of doing mental addition reasonably quickly, but some specific series of numbers (like the Fibonacci series, which I taught a way to add last December) can be added much quicker. So, we are going to need some more fastidiousness when we get our numbers to add.

The series we will be adding is the powers of two. Have your volunteer choose a number and list out that many powers of two. Let's say they say nine.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 =

Dont retrogress yet. All you have to do is double the final number in the sequence and then subtract one.

256 x 2 - 1

512 - 1

511

And there is your answer. If you have a little more prowess, you could try going up to sixteen numbers.

1 + 2 + 4 + 8 + 16 + 32 + 64 + 128 + 256 + 512 + 1024 + 2048 + 4096 + 8192 + 16384 + 32768 =

If you do the math, you will find that the answer is 65535.

This method is actually fairly easy to prove. Remember back when we did infinite series? Let's write that one we are subtracting as an infinite series.

2^0 + 2^1 + 2^2 + ... + 2^n-2 + 2^n-1 = 2^n - (2^-1 + 2^-2 + 2^-3 + ...)

Now, add that series to both sides. If we write it as an infinite series, we get:

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

The left side is already simplified at 2^n. How do we simplify the right side? I may have mentioned in a earlier post (I'm not sure which) how to simplify this type of series. Let me explain it again.

1. Find what number you multiply each number in the sequence by to get the next number in the sequence. Call that s.

2. Call the first and biggest number in the sequence x.

3. Use the formula x/(1-s) to get the total.

In this sequence, we are multiplying by 1/2. 1/2(2^n-1) can be written as 2^n-2.

The biggest number is the 2^n-1. So, we plug it into the formula to get:

(2^n-1)/(1 - 1//2)

(2^n-1)/(1/2)

2(2^n-1)

Using the law of exponents, we can simplify that to get:

2(2^n-1)

2^n-1+1

2^n

And there is the proof. Though the proof is kind of cool, I found it cool that you could add this many numbers up so quickly. Good luck fooling people with this one!

Now, add that series to both sides. If we write it as an infinite series, we get:

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

The left side is already simplified at 2^n. How do we simplify the right side? I may have mentioned in a earlier post (I'm not sure which) how to simplify this type of series. Let me explain it again.

1. Find what number you multiply each number in the sequence by to get the next number in the sequence. Call that s.

2. Call the first and biggest number in the sequence x.

3. Use the formula x/(1-s) to get the total.

In this sequence, we are multiplying by 1/2. 1/2(2^n-1) can be written as 2^n-2.

The biggest number is the 2^n-1. So, we plug it into the formula to get:

(2^n-1)/(1 - 1//2)

(2^n-1)/(1/2)

2(2^n-1)

Using the law of exponents, we can simplify that to get:

2(2^n-1)

2^n-1+1

2^n

And there is the proof. Though the proof is kind of cool, I found it cool that you could add this many numbers up so quickly. Good luck fooling people with this one!

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