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it goes on until nCn

p.s. what courses are you studying?

ive just graduated from year 12

neat.....which college/university/institute are you trying for

i think ill get into one of the iits

no problem

so r u indian?

wt are you doing??

i dont see an answer

hes right...that druvidfae is bloody bright

z^3 = 1 , how does that help ?

because all other binomial coefficients like nc1 and nc2 and so on have a (1 + w + w^2) multiplied

get it??

ok so ( 1 + z^1/3 ) ( 1 + z^1/3)( 1 + z ^1/3)

but roots of unity are not real

yes..first put in 1, then w and then w^2..u dont have to use only real nos ;)

one sec, im a bit rusty here,
( 1 + x ) ^m = 1 + mx + m(m-1)/2 x^2 + ..

same way if i wanted nC0 + nC5 + nC10... i would have put in the fifth roots of unity..

sorry had to leave for a moment

hey druvidfae wtr u studying>?

but n is not given?

the answer is in terms of n

oh wait, so youre using the fact that ( 1 + w + w^2 ) = ...

where w^3 = 1

= 0

1 + n + n(n-1)/2 *1 +....
1 + nw + n(n-1)/2 *w + ...
1 + nw^2 + (n)(n-1)/2 w^2 + ...

woops

1 + n + n(n-1)/2 *1 +....
1 + nw + n(n-1)/2 *w^2 + ...
1 + nw^2 + (n)(n-1)/2 w^4 + ..

ok, and w^4 = w , that sort of thing

whats the left side of this expression?

do you have an equation ?

i get 3 + 0 + 0 + ...

canto:yes... the principle can be extended to other problems as well.

whoa whoa , slow down there buddy

first thing is first, youre using the identity
( 1 + 1) ^n = Sum nCk , from k=0 to n ?

you guys leave out too many steps.

youre doing ( 1 + w)^n, (1+1)^n, and ( 1 + w^2) ^n, adding them

for this question we don't need the 2^n identity

right

yes

( 1 + w)^n + (1+1)^n, + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + ...

IS THAT WHAT YOU ARE DOING ????

and then stuff vanishes on the right side, but the left side i dont know what it is

the left side is a closed formula?

i would like you guys to actually solve it, lol

im still learning this stuff, sorry

i messed up

ok , im a little worried about the right side though

= 3 + 0 + 0 + nC3 + 0 + 0

why all non-3 get reduced to 0 .
that is the sequence we want

so the final answer is
[2^n + (1+w)^n + (1+w^2)^n ]/3 = nC0 + n C 3 + ... n C n

although n is not necessarily a multiple of 3

why all non-3 get reduced to 0 .
that is the sequence we want

im just checking , do you agree with my closed form formula ?

also we have to assume that n | 3

errr, 3 | n

hmmm, no that doesnt make sense, it should be less than 2^n

since nC0 + nC1 + ... n Cn = 2^n

oh you mean before dividing by 3 ?

we have deletion of terms

oh right

then you would zeroes as the last 1 or two terms ?

ok im going to test this , offhand i dont know the third root of unity of 1

e^ ( i * pi/3 ) ?

1 = e^(2*Pi *i) , so 1^(1/3) = e^(2pi/3 *i)

yes your forms are also correct

sorry, i have to start from fundamental principles, i have horrible memory

1 = e^( i*[ 2*Pi + 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3 +2pi/3*n)

so 1^1/3 = e ^ ( i * (2pi/3 , 4pi/3

oh its easier if we start with theta = 0

1 = e^( i*[ 0+ 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3*n)

so we have z^(1/3) = { 1 , e^(i*2pi/3) , e^ ( i*4pi/3)

i assume you have it memorized down well :) so i agree with that

so w and w^2 are conjugates, interesting

but it might be more succinct calculator wise to use exponential polar form

[2^n + (1+e^(i*2pi/3))^n + (1+e^(i*4pi/3))^n ]/3 = nC0 + n C 3 + ... n C n

now test, let n = 8

sorry my early values have a slight correction the real part should be negative.

let n = 8 , so we have
[2^8 + (1+e^(i*2pi/3))^8 + (1+e^(i*4pi/3))^8 ]/3 = 8C0 + 8 C 3 + ... 8C8 ?

or last term should be 8 C 6

yes

i get 85 for left side
85 + 3.3666666666666666666666E -13i

the last term should be\[(n & 6)\]

yessssssssssss

it works

*nc6

if n = 8

8C0 + 8C 3 + 8 C 6 = 85

so the imaginary part vanishes there

haha thats funny, my calcualto has 3.33333333336 E -13 i

so that vanishes , its like 3.3 *10^-13 * i

for the left side its
85 + 3.3666666666666666666666 E -13i
E means 10^

which calculator do you have
??

why didnt him actually solve it, thats annoying

TI 84

its easy to use, and fast to learn , the TI 89 is more powerful, but harder to use

im pissed, maple cant solve tan (105)

its an exact angle

TI 89 has multiple color codes, so light blue, light green for functions

my calculator:casio fx82 and fx 100
my numerical software :octave

it should be nC0 + nC3 + ... n C n divisible by 3

or something like n C k where k mod 3 is not equal to zero

not necessarily (due to the additional 2^n factor)

sum nC k , where nC k = 0 when k mod 3 ! = 0, and otherwise it is normal

i mean for the right hand side

yeah on the r.h.s.

great, i feel like i solved it. but you take credit for starting it

but we didnt simplify it all the way

LHS we have 2^n / 3 + `

oh right, youre real part should be negative

2pi/3 angle is in quadrant 2

also x^n = 1 gives you the pentagonal shapes

or z^n = 1 , i should say

northeast quadrant

northwest quadrant

starting from (1,0) angle 0 , going counterclockwise

ahhh.. memory so fickle art thou

quandrant 1 is 0 < theta < 90
q2 90 < theta < 180 , etc

q3 180 < theta < 270 , et

oh every nice, the 1-1 cancels the inside of the binomial

2^n / 3 + ( i * sqrt 3 / 2 )^n + (-i sqrt 3 / 2 ) ^n = n C0 + n C 3 + ..

notice that when you have n mod 3 = 0, then i^n + (-i)^n both go to zero

no i take that back

i^n + (-i)^n is interesting

it has the sequence for n = 1,2,3,
0 , -2, 0 , 2, 0, -2 , ...

oh thats because i^n +(- i) ^n = i^n ( 1 + (-1)^n)

i mesed up , i didnt see the 1 + (-1/2 + i sqrt 3 / 2 ) . so its

we have 2^n/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

2^n*1/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

and we can use an identity ( a + bi ) ^n + ( a - bi)^n

or z^n + z*^n = ...

where z* is the conjugate of z

i hope im not boring you

no definately not

i might need some help here

i did the case for i^n + (-i)^n

sure...and what came up

i^n ( 1 - (-1)^n ) , well a sequence

for n = 1 ,2,3
we have 0,-2,0,2,0,-2 ...

lets try the case a + bi ^n + ( a -bi)^n

yes that's what we're supposed to get remember with i the gap between terms is 2

oh the (a + bi)^n + ( a - bi)^n seems to diverge

i bet you by looking at the binomial form, we can simplify , by looking at the series

so its easier if we divide out the a ,

so ( 1 + ki ) ^n + ( 1 - ki)^n ,

oh very nice we get

1 + ki + (ki)^2 + k^3 *i^3 + ...
1 -ki + (ki^2) - k^3i^3 + ...

= 1 + (ki)^2 + (ki)^4 + ...

and we know 1 + x^2 + x^4 + ... = sum x^(2n)

so we get sum (ki)^(2n) for n=0...

that is equal to 1 / ( 1- (ki)^2 )

( 1 + ki ) ^n + ( 1 - ki)^n , = 1 / ( 1 - (ki)^2 ) ?

only if |ki|< 1

nevermind, this wont help us , lol

also i messed up, it should be
( 1 + ki ) ^n + ( 1 - ki)^n = 2 + 2(ki)^2 + 2 (ki)^4 + ...

oh crap, i left out the binomial coefficients

1 + nC1*ki + nC2(ki)^2 +nC3 k^3 *i^3 + ...
1 - nC0* ki + nC2* (ki^2) -nC3 k^3i^3 + ...

actually the power series of complex variables shows some diverging oscillation type things

oh its an alternating series

(( 1 + ki ) ^n + ( 1 - ki)^n) /2 = 1 - nC2 k^2 + nC4 k^4 - nC6 k^6 + ...

yay it works

i test it on k = 3, n = 4

i dont know if this is .... useful formula. can we simplify the right side?

might be a well known series

so what were you saying earlier, how can we simplify this result, the conjugates w, and w^2 appear

the final answer is then [ 2^n + ( 1/2 - i sqrt (3 )/ 2 ) ^n + ( 1/2+ i sqrt (3) / 2 ) ^n ] / 3

yes

but if we can find a nice simplification of ( a + bi)^n + ( a - bi)^n, that would be nice

and this argument works for any prime p , because ( 1 + w + w^2 + ... w^p-1) = 0

sorry but i dont see where the application is for this type of problem,

but its very pretty nonetheless