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Simplify (i+1)^{3200}-(i-1)^{3200}

waffles  Oct 29, 2017
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 #1
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Simplify (i+1)^{3200}-(i-1)^{3200}

 

= 2^1600  -   2^1600 =0  {Per Mathematica 11] !!!!.

Guest Oct 29, 2017
 #2
avatar+78538 
+1

(i + 1)^3200  =

 

i^3200 + ai^3199 + bi^3198  + ci^3197 + di^3196 +  .....+ di^4 + ci^3 + bi ^2 + ai + 1

 

 

(i - 1)^3200  = 

 

i^3200 - ai^3199 + bi^3198  - ci^3197 + di^3196 -  .....+ di^4 - ci^3 + bi ^2 - ai + 1

 

So

 

(i + 1)^3200  - (i - 1)^3200    leaves

 

2 [ ai^3199 +  ci^3197 + ......+ ci^3 + ai  ]   =

 

2  [ a (-i + i)  + c (i + - i)  +  e( -i + i) + g ( i + - i)  +  ....... ]  =  

 

2 [ a * 0  +  c * 0  +  e * 0  + g * 0  +  ......  ]  =

 

2 [ 0 ]  =

 

0

 

 

 

cool cool cool

CPhill  Oct 30, 2017
edited by CPhill  Oct 30, 2017
edited by CPhill  Oct 30, 2017
 #3
avatar+18712 
+1

Simplify (i+1)^{3200}-(i-1)^{3200}

 

\(\begin{array}{|rclrcl|} \hline && \mathbf{(i+1)^{3200}-(i-1)^{3200}} \\ &=& (i+1)^{2\cdot 1600}-(i-1)^{2\cdot 1600} \\ &=& \left( (i+1)^{2} \right)^{1600}- \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i+1)^{2} &=& i^2+2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1+2i+1 \\ && \quad & \quad &=& 2i \\ &=& (2i)^{1600} - \left( (i-1)^{2} \right)^{1600} \quad & |\quad (i-1)^{2} &=& i^2-2i+1 \qquad i^2 = -1\\ && \quad & \quad &=& -1-2i+1 \\ && \quad & \quad &=& -2i \\ &=& (2i)^{1600} - (-2i)^{1600} \\ &=& (2i)^{1600} - (-1)^{1600}(2i)^{1600} \quad & \quad (-1)^{1600} = 1 \\ &=& (2i)^{1600} - (2i)^{1600} \\ &\mathbf{=}& \mathbf{0} \\ \hline \end{array}\)

 

laugh

heureka  Oct 30, 2017
edited by heureka  Oct 30, 2017

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