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How do you solve

i100+i99+i98++i1+i0+i1++i99+i100

 Jun 11, 2018
 #1
avatar+985 
+3

i100+i99+i98++i1+i0+i1++i99+i100

 

i=1, we start simple and try to find a pattern.

 

i0=1i1=ii2=1i3=ii4=1i5=ii6=1i7=i

Every 4 terms, the sequence rotates, 1,i,1,i

 

Their sum is: 1+i1i=0

 

Since every four terms cancel out, we just need to know what term the sequence starts and ends. 

 

i100=1i100=1

i100+i99+i98++i1+i0+i1++i99+i100=1+i1i+1++1+i1i+1=1

 

I hope this helped,

 

Gavin

 Jun 11, 2018
edited by GYanggg  Jun 11, 2018
 #2
avatar+26396 
+2

How do you solve
i100+i99+i98++i1+i0+i1++i99+i100

 

i100+i99+i98+i97++i2+i1+i0+i1+i2+i3+i4+i99+i100=i100(1+i)+i98(1+i)++i2(1+i)+i0+i1(1+i)+i3(1+i)++i99(1+i)1+i=11=0=i100×0+i98×0++i2×0+i0+i1×0+i3×0++i99×0=0+i0+0=i0=1

 

laugh

 Jun 12, 2018
edited by heureka  Jun 12, 2018

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