Expression for sum of numbers

qeb

Feb 2016
5
0
London
1: 3,5: 7,9,11 ....sum of these groups can be expressed as n raised to 3. How can I prove this?
 

Plato

MHF Helper
Aug 2006
22,490
8,653
1: 3,5: 7,9,11 ....sum of these groups can be expressed as n raised to 3. How can I prove this?
$n^3:~1,~8,~27,~64,~\cdots$ add numbers the given groups.

Is the next group $13,~15,~17,~19~?$
 

Debsta

MHF Helper
Oct 2009
1,343
620
Brisbane
Interesting.
Group 1 : 1 Sum 1 : 1
Group 2 : 3,5 Sum 2 : 8
Group 3 : 7, 9,11 Sum 3 : 27
Group 4 : 13,15,17,19 Sum 4 : 64
etc
Each group is an arithmetic progression.
If the group number is x, then the first number in the group is x(x-1)+1
Use the sum of an arithmetic progression formula with a=x(x-1)+1, d=2 and n = x (seeing that the group number is the same as the number of terms in the group)
Then do some algebra. Falls out nicely!

(Not sure if this constitutes a "real" proof as a=x(x-1)+1 was found by observation.)
 
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Dec 2013
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Colombia
Nicely done.
(Not sure if this constitutes a "real" proof as a=x(x-1)+1 was found by observation.)
Note that the first \(\displaystyle n\) groups use \(\displaystyle \sum \limits_{k=1}^n k\) odd numbers. So the first number of each group is the \(\displaystyle \left(1+\sum \limits_{k=1}^n k\right)\)th odd number. You may well already know the formula for the sum of the first \(\displaystyle n\) natural numbers (1, 2, 3, ...). this will give your value for \(\displaystyle a\).