IIT JEE , AIEEE Forum - Mathematics , Algebra

rtiit

If a (not equal to 1) is the nth root of unity then find the sum of the following series:

1+3a+5a^2+........................upto n terms

rishipratimm

i'm nt exactly sure....but is d ans 0???

rtiit

bt the problem is that i dont know the ans

can u give the detailed solution??

rishipratimm

i'm nt exactly sure...
there's a rule dat for n roots of unity,sum of all d roots is always 0...
like 1+omega+omega^2=0....
mayb u hv to use it here....
nt sure bout d soln...
sorry bout dat....

raulrag009

Look at my next post

i had accidently submitted twice

raulrag009

Here is my soln

$a^{n}=1\\\\ S_{n}=1+3a+5a^{2}...............(2n-1)a^{n-1}\\\\ aS_{n}=0+a+3a^{2}................(2n-1)a^{n}\\\\ Subtract\; both\\\\ S_{n}(1-a)=a+\frac{2a(1-a^{n-1})}{(1-a)^{2}}-(2n-1)a^{n}\\\\ As\quad\;a^{n}=1\\ a^{(n-1)}=a\\\\ S_{n}=\frac{a}{1-a}+\frac{2a(1-a)}{(1-a)^{2}}-\frac{(2n-1)}{1-a}\\\\ S_{n}=\frac{3a-2n+1}{1-a}$

hsbhatt

Hint:This is an AGP

Let S = 1+3a+5a²+...+(2n-1)a^n-1

aS = a+3a²+.....+(2n-3)a^n-1 + (2n-1)aⁿ

So S(1-a) = 1+2(a+a²+a³+...+a^n-1) - (2n-1)aⁿ

Now aⁿ-1=0

1+a+a²+...+a^n-1 = 0 as a

1

PS: sorry raul, someone had asked for help. didnt notice your post

edit: the exponent has been corrected after raul pointed it out

raulrag009

No problem sir , but i think it should hav been a to the power (n-1) in the sequence

sourab_MCA

let S=1+3a+5a^2+..........nth term
> aS= a+3a^2+.............n-1 term+a*nth term
(subtracting)
S(1-a)=1+(2a+2a^2+.........n-1)-a*nth term
=1+2a(1+a^2+.........n-1)-a*nth term
= 1+2a(0)-a*nth term from property of nth root of unity
S=1-(a*nth term)/1-a

find the nth term of the series and proceed........

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