Loading [MathJax]/jax/output/SVG/jax.js
 
+0  
 
0
182
2
avatar+96 

Compute i+i^2+i^3.............i^258+i^259

 Jan 15, 2023
 #2
avatar+14 
0

Because (i,i2,i3,i4)=(i,1,i,1), we have  i+i2+i3+i258+i259=64(i1i+1)+i257+i258+i259. So, the desired sum is i257+i258+i259=i1i=1.

 Jan 16, 2023

1 Online Users