1, 8, 27, 64, ... What are the next 3 numbers in the sequence and how do I figure them out?
It's a pattern...... I guess I'd suggest just looking at it a bit longer and giving it some thought.
every term is x^3 You have: 1^3, 2^3, 3^3, 4^3, ... So, the next 3 would be: 5^3, 6^3, 7^3 or 125, 216, 343
All good now @xbrittxcheerx92x ?
I still don't really understand how you did this problem :(
The key is starting by associating the number 1 with the first term, the number 2 with the second term, 3 with the 3rd, 4 with the 4th and seeing if for any of them (I'd look at terms 2,3,4) have a relationship with the number of the term and its value. for instance, trying to find a relationship between 4 and 64 or between 3 and 27. If you look at 2 of them together like this, you'll see that they are cubes.
Because 3 x 3 x 3 = 27 and 4 x 4 x 4 = 64 and 2 x 2 x 2 = 8 same for 1
So each term in the sequence is just that number cubed.
It helps if you write the position under each number in the sequence: 1 8 27 64 1 2 3 4 That's an effective visual aid.
Oh! ok. I get it now. Thanks :)
So, the end result: 1 8 27 64 125 216 343 1^3 2^3 3^3 4^3 5^3 6^3 7^3 1 2 3 4 5 6 7
If I graphed y = x^3, would you like to see that? That also is a good visual aid. @xbrittxcheerx92x
This can help:
Thank you so much! Can you help me out with another equation please?
ok! Don't forget to medal!
1, 12, 123, 1234, 12345, ...
That one is a quirky type of sequence. You just put the next number after the end of each number. Notice how the 3rd term has a "3" put after the "12"
And then for the 4th term, a "4" is put after the "123" which is the 3rd term. It's like building blocks or stacking.
Mathematically, you could say the rule to get a term is just to take the previous term, multiply it by 10, and then add the positional number of the term to that product. For instance: The 7th term is 10 times the 6th term plus 7 or 123456 x 10 + 7
Making sense now @xbrittxcheerx92x ?
yes. Thank you so much!
Good luck in all of your studies and thx for the recognition! @xbrittxcheerx92x
No problem. Thank you! :)
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