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Require Help With Simplification
- Thread starter csm
- Start date
Clarify please. Like, (-1)^2n+2 means what?
(-1)^2 times (n+2)?
(-1)^(2n) + 2 ?
(-1)^(2n + 2) ?
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Hi Dennis, yes. Its (-1) to the power of 2(n+1).
D
Deleted member 4993
Guest
How are these steps possible?
[(-1)^(2n+1)] . [(2n)^2] + [(-1)^(2n+2)] . [(2n+1)^2]
= [(-1) . (4n^2) + (4n^2 +4n + 1)
= 4n + 1
correct...
Please use * for multiplication - at least on this board. It avoids confusion with decimal point.
@subhotoshkhan
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could you please explain how the steps proceed each other?
correct...
Please use * for multiplication - at least on this board. It avoids confusion with decimal point.
___________________
could you please explain how the steps proceed each other?
Assuming n is an integer, this is really a problem about odd and even powers of - 1. Odd powers = - 1, and even powers = + 1. Watch below how it works using the laws of exponents.How are these steps possible?
[(-1)^(2n+1)] . [(2n)^2] + [(-1)^(2n+2)] . [(2n+1)^2]
= [(-1) . (4n^2) + (4n^2 +4n + 1)
= 4n + 1
\(\displaystyle (- 1)^{(2n + 1)} = (-1)^{(2n)} * (-1)^1 = \left(\{(-1)^2\}^n\right) * (-1)^1 = (1)^n * (- 1) = 1 * (- 1) = - 1.\) You OK with that?
\(\displaystyle (- 1)^{(2n + 2)} = \left\{(-1)^2\right\}^{(n + 1)} = 1^{(n + 1)} = 1.\) Is that clear?
\(\displaystyle So\ \left\{(-1)^{2n + 1)} * (2n)^2\right\} + \left\{(-1)^{(2n + 2)} * (2n + 1)^2\right\} =\)
\(\displaystyle \left\{(-1) * (2n)^2\right\} + \left\{1 * (2n + 1)^2\right\} =\)
\(\displaystyle \left\{(-1) * 4n^2\right\} + \left\{1 * (4n^2 + 4n + 1)\right\} =\)
\(\displaystyle - 4n^2 + 4n^2 + 4n + 1 =\)
\(\displaystyle 4n + 1\)
Clear now?