SOLUTION: 1+2+3=40 2+3+4=85 3+4+5=148 4+5+6=229 5+6+7=???

Question 991899: 1+2+3=40
2+3+4=85
3+4+5=148
4+5+6=229
5+6+7=???
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
Ask somewhere in the site on puzzles,
not in the algebra.com

By the way, this system has 5 equations and 7 unknowns.
So you can choose 1 and 2 by the arbitrary way,
then to determine 3,
then to determine 4,
and so on step by step . . . till get 7.

The solution is not unique. It depends on two arbitrary constants.

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