23is d value 0 ???
a is cube root of unity na ???
if a+\frac{1}{a}=-1 find the value of (a^2+a+1)+(a^4+a^2+1)+(a^6+a^3+1)+(a^8+a^4+1)+...+(a^{2006}+a^{1003}+1)
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6 Answers
21put w (which is cube root of unity) in place of a
(a2 +a4 +a6............a2006) +(a +a2+a3...............a1003) + 1003
a2(1-(a2)1003(1-a2)) +a(1-a1003)(1-a) +1003
a2(1-a2006)1-a2 +a(1-a1003)(1-a) +1003
putiing w in place of a
since w2006 = w2
w1003=w
we get ......
=w2 +w +1003
since 1+w+w2 =0
=1+w+w2 +1002
=1002
1thanks Deepak but i couldnt understand your solution as i havent done cube root of unity as yet
62fibonacci..
cube root of infinity are the solution for x3=1
whci is same as (x-1)(x2+x+1) =0
two roots are complex and they are generally denoted as w and w2 (read omega)
Also, we can prove that w3n+1+(w2)3n+1+1 = 0
and that w3n+2+(w2)3n+2+1 = 0
and that w3n+(w2)3n+1 = 3
The last one is the most trivial..
I hope this helps you :)