nc0 + nc3 + nc6 ......Please help

- anonymous

nc0 + nc3 + nc6 ......Please help

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- anonymous

it goes on until nCn

- anonymous

expand (1+ x)^n bionomialy then put x = the cube roots of unity you'll get 3 equations which on adding and subtracting will allow you to create the series you want

- anonymous

p.s. what courses are you studying?

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## More answers

- anonymous

ive just graduated from year 12

- anonymous

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

- anonymous

i think ill get into one of the iits

- anonymous

no problem

- anonymous

so r u indian?

- anonymous

wt are you doing??

- anonymous

i dont see an answer

- anonymous

hes right...that druvidfae is bloody bright

- anonymous

z^3 = 1 , how does that help ?

- anonymous

u put in the cube roots of unity one by one and add all three equations...the required series comes out

- anonymous

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

- anonymous

get it??

- anonymous

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

- anonymous

but roots of unity are not real

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

sorry had to leave for a moment

- anonymous

hey druvidfae wtr u studying>?

- anonymous

but n is not given?

- anonymous

the answer is in terms of n

- anonymous

yes you would the problem...is that you would have lrage number of equations also in other situations only substitutions for a prime no. will work.

- anonymous

Though I'm doing my degree in computer science I am interested in all branches of science and mathematics ... particular abstract maths(algebra,m-th,etc)

- anonymous

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

- anonymous

where w^3 = 1

- anonymous

= 0

- anonymous

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

- anonymous

woops

- anonymous

- anonymous

since the roots repeat themselves modularly i.e.after every 3 powers. the sequence effectively has only three variable inserting subsequent values allows the offset to change thus allowing us to affect every third element symmetrically

- anonymous

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

- anonymous

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

- anonymous

whats the left side of this expression?

- anonymous

do you have an equation ?

- anonymous

i get 3 + 0 + 0 + ...

- anonymous

its similar to interference in physics .Waves with a particular offset (phi) affect the the superposition in a certain way (standing waves, packets ,etc) here we are changing the wavelength (period) rather than the offset to match our sequence.once we've got our wavelength we continue to change the offset to get the shape we need.

- anonymous

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

- anonymous

whoa whoa , slow down there buddy

- anonymous

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

- anonymous

you guys leave out too many steps.

- anonymous

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

- anonymous

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

- anonymous

right

- anonymous

yes

- anonymous

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

- anonymous

( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + nC3 ( 1 + w^3 + w^8 ) + ... nCn ( 1 + w^n + w^(2^n))

- anonymous

IS THAT WHAT YOU ARE DOING ????

- anonymous

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

- anonymous

the left side is a closed formula?

- anonymous

i would like you guys to actually solve it, lol

- anonymous

im still learning this stuff, sorry

- anonymous

i messed up

- anonymous

( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + nC3 ( 1 + w^3 + w^6 ) + ... nCn ( 1 + w^n + w^(2n))

- anonymous

yes...as for the left side notice that the w and w^2 factors have imaginary parts which are inverse of one another expanding this using euler's form will reveal that the imaginary parts get canceled out.

- anonymous

ok , im a little worried about the right side though

- anonymous

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

- anonymous

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

- anonymous

( 1 + w)^n + (1+1)^n + ( 1 + w^2) ^n = 3 + 3*n( 1 + w + w^2) + 3*nC2 ( 1 + w^2 + w^4 ) + 3*nC3 ( 1 + w^3 + w^6 ) + ... 3*nCn ( 1 + w^n + w^(2n))

- anonymous

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

- anonymous

although n is not necessarily a multiple of 3

- anonymous

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

- anonymous

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

- anonymous

also we have to assume that n | 3

- anonymous

errr, 3 | n

- anonymous

yeah the l.h.s. reduces to something like 3^n (note there are probably some more factors i'm missing)

- anonymous

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

- anonymous

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

- anonymous

oh you mean before dividing by 3 ?

- anonymous

if n is not a multiple of the then the last 1 or 2 factors do not appear in the final series........since series is up to n.

- anonymous

we have deletion of terms

- anonymous

oh right

- anonymous

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

- anonymous

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

- anonymous

e^ ( i * pi/3 ) ?

- anonymous

no... \[w =(1+ i \sqrt{3})/2 ; w^2 = (1-i \sqrt{3})/2\] expand both bionomially notice that odd powers cancel and the remaining terms are of the form 3^n

- anonymous

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

- anonymous

yes your forms are also correct

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

oh its easier if we start with theta = 0

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

so w and w^2 are conjugates, interesting

- anonymous

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

- anonymous

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

- anonymous

now test, let n = 8

- anonymous

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

- anonymous

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 ?

- anonymous

or last term should be 8 C 6

- anonymous

does your calculator allow complex calculations......i.e. calculations involving iota based calculations.

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

yessssssssssss

- anonymous

it works

- anonymous

*nc6

- anonymous

if n = 8

- anonymous

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

- anonymous

so the imaginary part vanishes there

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

which calculator do you have
??

- anonymous

why didnt him actually solve it, thats annoying

- anonymous

TI 84

- anonymous

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

- anonymous

im pissed, maple cant solve tan (105)

- anonymous

its an exact angle

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

i mean for the right hand side

- anonymous

yeah on the r.h.s.

- anonymous

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

- anonymous

but we didnt simplify it all the way

- anonymous

LHS we have 2^n / 3 + `

- anonymous

oh right, youre real part should be negative

- anonymous

2pi/3 angle is in quadrant 2

- anonymous

also x^n = 1 gives you the pentagonal shapes

- anonymous

or z^n = 1 , i should say

- anonymous

northeast quadrant

- anonymous

northwest quadrant

- anonymous

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

- anonymous

ahhh.. memory so fickle art thou

- anonymous

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

- anonymous

q3 180 < theta < 270 , et

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

no i take that back

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

where z* is the conjugate of z

- anonymous

i hope im not boring you

- anonymous

no definately not

- anonymous

i might need some help here

- anonymous

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

- anonymous

sure...and what came up

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

so its easier if we divide out the a ,

- anonymous

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

- anonymous

oh very nice we get

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

only if |ki|< 1

- anonymous

nevermind, this wont help us , lol

- anonymous

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

- anonymous

oh crap, i left out the binomial coefficients

- anonymous

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

- anonymous

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

- anonymous

oh its an alternating series

- anonymous

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

- anonymous

yay it works

- anonymous

i test it on k = 3, n = 4

- anonymous

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

- anonymous

might be a well known series

- anonymous

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

- anonymous

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

- anonymous

yes

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

but its very pretty nonetheless

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