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

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

Mathematics
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it goes on until nCn
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
p.s. what courses are you studying?

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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 ?
u put in the cube roots of unity one by one and add all three equations...the required series comes out
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
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.
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)
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
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)
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
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 + ...
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.
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 ) + ...
( 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))
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
( 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))
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.
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
( 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))
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
yeah the l.h.s. reduces to something like 3^n (note there are probably some more factors i'm missing)
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 ?
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.
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 ) ?
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
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
does your calculator allow complex calculations......i.e. calculations involving iota based calculations.
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

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