+0

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

0
56
3
+352

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

waffles  Oct 29, 2017
Sort:

#1
0

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

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

Guest Oct 29, 2017
#2
+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

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

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

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

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

### 25 Online Users

We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details