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Please find the next 4 terms in the following sequence and thanks:

1, 4, 9, 16, 25, 36, 49, 64, 81, 10, 22............?

 Jan 8, 2016
 #1
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1^0   2^2  3^2  4^2   5^2  6^2  7^2  8^2  9^2 

10     22    32   42    52    62    72                      ?????

 Jan 9, 2016
 #2
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After thinking about it for a few minutes, it suddenly hit me!!!!! I think it goes likes this:

"The sum of the digits of each term X number of term=the term!". So, we have:

1 X 1=1 first term, 2 X 2=4 second term, 3 X 3=9 third term..............

10th term=(1+0) X 10=10th term, 11th term=(1+1) X 11=22 , which is the 11th term, 12th term=(1+2) X 12=36, which is the 12th term, 13th term=(1+3) X 13=52, which is the 13th term, 14th term=(1+4) X 14=70............and so on.....90, 112, 136.............etc. So, the sequence is:

1, 4, 9, 16, 25, 36, 49, 64, 81, 10, 22, 36, 52, 70, 90, 112, 136...........etc.(I think!!!!!????).

 Jan 9, 2016
 #3
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Very nice, guest......I would have never seen that......!!!!!!

 

 

 

cool cool cool

 Jan 9, 2016
 #4
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Thank you CPhill, unless it gets shot down by the poster!!.

 Jan 9, 2016
 #5
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Thank you Guest #2. That is exactly what it says in my book but couldn't figure out how it went.

 Jan 9, 2016

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