The 2011 power of (- 2) + (- 2) to the 2010 power + (- 2) to the 2009 power + ··· + (- 2) to the 2nd power + (- 2) + 1

The 2011 power of (- 2) + (- 2) to the 2010 power + (- 2) to the 2009 power + ··· + (- 2) to the 2nd power + (- 2) + 1

The 2011 power of (- 2) + (- 2) to the 2010 power + (- 2) to the 2009 power + ··· + (- 2) to the 2nd power + (- 2) + 1
=1×[1-(-2)^2012]/(1+2)
=[1-(-2)^2012]/3
This is my conclusion after meditation,
If you can't ask, I will try my best to help you solve it~
If you are dissatisfied and willing, please understand~

How to calculate the square of a plus the negative quadratic power of a

=- 2 times of a (4 times of a + 1)

Calculate the (n + 1) square of 2 * (2n + 1) power of (1 / 2) / N power of 4 * the negative 2 power of 8

I think there's something wrong with the title
The original formula = (2 + 1) (the 2nd power of 2 + 1) (the 4th power of 2 + 1) (2n power of 2 + 1)
So, let's use the square difference formula
(2-1) * (2 + 1) (2's 2nd power + 1) (2's 4th power + 1) (2n power of 2 + 1)
=(2nd power of 2-1)) (2nd power of 2 + 1) (4th power of 2 + 1) (2n power of 2 + 1)
=.
=(2n power of 2-1) (2n power of 2 + 1)
=2 to the power of 2 (n + 1) - 1
Then the original formula = [2 (n + 1) power-1] / (2-1) = 2's 2 (n + 1) power-1

1 / 3 to the 1000 power × 3 to the negative 1000 power =

The minus 2000 power of 3

Simplify (1 + Tan ^ 2 a) cos ^ 2 A

=（1+sin^2 a/cos^2 a）cos^2 a
=sin^2 a+cos^2 a
=1

Simplify cos (2 + α) Tan (PAI + α) / cos (Pie / 2 + α)

According to the induction formula
Cos (2 + α) Tan (PAI + α) / cos (Pie / 2 + α)
=cos(α)tan( α)/(-sinα)
=(cosα*sinα/cosα)/(-sinα)
=sinα/(-sinα)
=-1

Simplified calculation of cos π / 3-tan π / 4 + cos π - Sin 3 π / 2

=1/2-1-1-(-1)
=-1/2

Simplification of COS (α + π) Tan (2 π + α)

cos(α+π)tan(2π+α)=-cosa*tana=-sina

It was proved that: (1 / sin α - sin (180 ° + α)) / (1 / cos (540 ° - α) + cos (360 ° - α)) = 1 / (Tan α) ^ 3

Is it: (- 1 / sin α - sin (180 ° + α)) / (1 / cos (540 ° - α) + cos (360 ° - α)) = 1 / (Tan α) ^ 3
prove:
Left = [- 1 / Sina - (- Sina)] / [1 / (- COSA) + cosa]
=[(-1+sin^2a)/sina]/[(cos^2-1)/cosa]
=-cos^2a/sina*cosa/(-sin^2a)
=cos^3a/sin^3a
=1/(tana)^3
=Right

.[1/cos(-α)+cos(180°+ α )]/[1/sin(540°-α)+sin(360°-α)]=tan^3α

Cos (- α) = cos α, cos (180 ° + α) = - cos α sin (540 ° - α) = sin α sin (360 ° - α) = - sin α