SOLUTION: I cannot find the solution to -5, 10, -15, 20, ....

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Question 110183This question is from textbook Algebra II Tennessee Edition
: I cannot find the solution to -5, 10, -15, 20, .... This question is from textbook Algebra II Tennessee Edition

Found 2 solutions by solver91311, scott8148:
Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!
First notice that the sign on each term alternates and the odd numbered terms are negative. That means a good place to start is to have each term contain the factor %28-1%29%5En, which yields a positive 1 for all even n, and a negative 1 for all odd n.
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The next thing to notice is that the absolute value of each element increases by 5 each subsequent element, so 5n is a good expression for that.
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Putting it all together you get:
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%28-1%29%5En%2A5n
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Check:
%28-1%29%5E1%2A5%2A1=%28-1%29%2A5=-5
%28-1%29%5E2%2A5%2A2=1%2A5%2A2=10
%28-1%29%5E3%2A5%2A3=%28-1%29%2A5%2A3=-15
%28-1%29%5E4%2A5%2A4=5%2A4=20
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And the 5th term should be:
%28-1%29%5E5%2A5%2A5=%28-1%29%2A5%2A5=-25
:
Exercise for the student: What would you do if your series had the odd terms positive and the even terms negative, i.e. the exact opposite of the problem given?

Answer by scott8148(6628) About Me  (Show Source):
You can put this solution on YOUR website!
(-1)^n will give the changing signs ... 5n will give the numbers
so 5n(-1)^n