1,1,2,4,16,128,4096

What are the next 3 terms.

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- May 4th 2010, 08:16 PMintegralNumber pattern.
1,1,2,4,16,128,4096

What are the next 3 terms. - May 4th 2010, 08:23 PMpickslides
the next $\displaystyle 2$ are $\displaystyle 1048576$ and $\displaystyle 8589934592$ I will let someone else reveal the third (Rofl)

- May 4th 2010, 08:36 PMWilmer
Next 3:

18014398509481984, 309485009821345068724781056, 11150372599265311570767859136324180752990208 - May 5th 2010, 03:39 AMintegral
I would have thought that would have taken some time. It would have for me :(.

Please tell me how you can find patterns so quick? Or is it just, your that good at pattern recognition? Thanks - May 5th 2010, 03:56 AMpickslides
This one was quite easy. This is what to do. Write the numbers down and start looking at any relationships between consecutive numbers.

In this case look at the ratios between each consecutive numbers. For the next term consider the relationship between the previous two ratios.

$\displaystyle 1 {\color{red}\times 1 = } , 1 {\color{red}\times 2 = } , 2 {\color{red}\times 2 = } , 4 {\color{red}\times 4 = } , 16 {\color{red}\times 8 = } , 128 {\color{red} \times 32 = } ,\dots $ - May 5th 2010, 03:58 AMundefined
- May 5th 2010, 03:42 PMintegral
Ohrly? Interesting.

I was playing around to find a number pattern so I started with pascals triangle.

$\displaystyle

\binom{n}{k}$

then I added up each row, but not all rows. I chose a row correlating to a financing number subtracted by 1.

so

$\displaystyle

\binom{(F_{n}+F_{n-1})-1}{k}$

which gives

$\displaystyle \binom{1-1}{k}$

$\displaystyle \binom{1-1}{k}$

$\displaystyle \binom{2-1}{k}$

$\displaystyle \binom{3-1}{k}$

ect.

But I wanted the sum so.

$\displaystyle \sum^{F_n-1}_{k}\binom{F_n-1}{k}$

$\displaystyle \left [ \sum^{0}_{k}\binom{0}{k} \right ]=1$

$\displaystyle \left [ \sum^{0}_{k}\binom{0}{k} \right ]=1$

$\displaystyle \left [\sum^{1}_{k}\binom{1}{k}\right ]=2$

$\displaystyle \left [ \sum^{2}_{k}\binom{2}{k}\right ]=4$

$\displaystyle \left [ \sum^{4}_{k}\binom{4}{k}\right ]=16$

$\displaystyle \left [ \sum^{7}_{k}\binom{7}{k}\right ]=128$

While your method works yes (although I did not think of it)

It does not explain the first term, only all after.

**1,1**,2,4,16

So, you're answers work but are technically incorrect.

:D

$\displaystyle

\int$ - May 5th 2010, 03:54 PMundefined
- May 5th 2010, 04:36 PMWilmer
- May 5th 2010, 04:47 PMintegral
I suppose, indeed.

Interesting non the less (Giggle) - May 7th 2010, 03:28 AMsimplependulum
Look at the power of the pattern

$\displaystyle 0,0,1,2,4,7,12 ....$

If we add one in each term

we have

$\displaystyle 1 , 1 , 2,3,5,8,13 ... $

Fibonacci sequence (Talking) - May 7th 2010, 05:00 PMintegral
- May 7th 2010, 07:52 PMGusbob
- May 8th 2010, 06:29 PMintegral
I see, so any sequence you place at the top of your binomial coefficient will be two to the power of that sequence?

:D awesome!