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anonymous
 5 years ago
nc0 + nc3 + nc6 ......Please help
anonymous
 5 years ago
nc0 + nc3 + nc6 ......Please help

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0expand (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
 5 years ago
Best ResponseYou've already chosen the best response.0p.s. what courses are you studying?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ive just graduated from year 12

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0neat.....which college/university/institute are you trying for

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i think ill get into one of the iits

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hes right...that druvidfae is bloody bright

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0z^3 = 1 , how does that help ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0u put in the cube roots of unity one by one and add all three equations...the required series comes out

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0because all other binomial coefficients like nc1 and nc2 and so on have a (1 + w + w^2) multiplied

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so ( 1 + z^1/3 ) ( 1 + z^1/3)( 1 + z ^1/3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but roots of unity are not real

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes..first put in 1, then w and then w^2..u dont have to use only real nos ;)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0one sec, im a bit rusty here, ( 1 + x ) ^m = 1 + mx + m(m1)/2 x^2 + ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0same way if i wanted nC0 + nC5 + nC10... i would have put in the fifth roots of unity..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry had to leave for a moment

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hey druvidfae wtr u studying>?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the answer is in terms of n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes 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
 5 years ago
Best ResponseYou've already chosen the best response.0Though I'm doing my degree in computer science I am interested in all branches of science and mathematics ... particular abstract maths(algebra,mth,etc)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh wait, so youre using the fact that ( 1 + w + w^2 ) = ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 + n + n(n1)/2 *1 +.... 1 + nw + n(n1)/2 *w + ... 1 + nw^2 + (n)(n1)/2 w^2 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0Though I'm doing my degree in computer science I am interested in all branches of science and mathematics ... particular abstract maths(algebra,mth,etc)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since 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
 5 years ago
Best ResponseYou've already chosen the best response.01 + n + n(n1)/2 *1 +.... 1 + nw + n(n1)/2 *w^2 + ... 1 + nw^2 + (n)(n1)/2 w^4 + ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok, and w^4 = w , that sort of thing

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whats the left side of this expression?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0do you have an equation ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 3 + 0 + 0 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its 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
 5 years ago
Best ResponseYou've already chosen the best response.0canto:yes... the principle can be extended to other problems as well.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0whoa whoa , slow down there buddy

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0first thing is first, youre using the identity ( 1 + 1) ^n = Sum nCk , from k=0 to n ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0you guys leave out too many steps.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0youre doing ( 1 + w)^n, (1+1)^n, and ( 1 + w^2) ^n, adding them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for this question we don't need the 2^n identity

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 1 + w)^n + (1+1)^n, + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 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
 5 years ago
Best ResponseYou've already chosen the best response.0IS THAT WHAT YOU ARE DOING ????

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and then stuff vanishes on the right side, but the left side i dont know what it is

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the left side is a closed formula?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i would like you guys to actually solve it, lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im still learning this stuff, sorry

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 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
 5 years ago
Best ResponseYou've already chosen the best response.0yes...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
 5 years ago
Best ResponseYou've already chosen the best response.0ok , im a little worried about the right side though

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= 3 + 0 + 0 + nC3 + 0 + 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why all non3 get reduced to 0 . that is the sequence we want

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 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
 5 years ago
Best ResponseYou've already chosen the best response.0so the final answer is [2^n + (1+w)^n + (1+w^2)^n ]/3 = nC0 + n C 3 + ... n C n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0although n is not necessarily a multiple of 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why all non3 get reduced to 0 . that is the sequence we want

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im just checking , do you agree with my closed form formula ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also we have to assume that n  3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yeah the l.h.s. reduces to something like 3^n (note there are probably some more factors i'm missing)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmmm, no that doesnt make sense, it should be less than 2^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0since nC0 + nC1 + ... n Cn = 2^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh you mean before dividing by 3 ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0if 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
 5 years ago
Best ResponseYou've already chosen the best response.0we have deletion of terms

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then you would zeroes as the last 1 or two terms ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok im going to test this , offhand i dont know the third root of unity of 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no... \[w =(1+ i \sqrt{3})/2 ; w^2 = (1i \sqrt{3})/2\] expand both bionomially notice that odd powers cancel and the remaining terms are of the form 3^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = e^(2*Pi *i) , so 1^(1/3) = e^(2pi/3 *i)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes your forms are also correct

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry, i have to start from fundamental principles, i have horrible memory

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = e^( i*[ 2*Pi + 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3 +2pi/3*n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so 1^1/3 = e ^ ( i * (2pi/3 , 4pi/3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh its easier if we start with theta = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 = e^( i*[ 0+ 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3*n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we have z^(1/3) = { 1 , e^(i*2pi/3) , e^ ( i*4pi/3)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i assume you have it memorized down well :) so i agree with that

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so w and w^2 are conjugates, interesting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but it might be more succinct calculator wise to use exponential polar form

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0[2^n + (1+e^(i*2pi/3))^n + (1+e^(i*4pi/3))^n ]/3 = nC0 + n C 3 + ... n C n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry my early values have a slight correction the real part should be negative.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let 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
 5 years ago
Best ResponseYou've already chosen the best response.0or last term should be 8 C 6

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0does your calculator allow complex calculations......i.e. calculations involving iota based calculations.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i get 85 for left side 85 + 3.3666666666666666666666E 13i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the last term should be\[(n & 6)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.08C0 + 8C 3 + 8 C 6 = 85

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the imaginary part vanishes there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0haha thats funny, my calcualto has 3.33333333336 E 13 i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so that vanishes , its like 3.3 *10^13 * i

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the left side its 85 + 3.3666666666666666666666 E 13i E means 10^

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0which calculator do you have ??

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0why didnt him actually solve it, thats annoying

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0its easy to use, and fast to learn , the TI 89 is more powerful, but harder to use

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0im pissed, maple cant solve tan (105)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0TI 89 has multiple color codes, so light blue, light green for functions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0my calculator:casio fx82 and fx 100 my numerical software :octave

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it should be nC0 + nC3 + ... n C n divisible by 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or something like n C k where k mod 3 is not equal to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not necessarily (due to the additional 2^n factor)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sum nC k , where nC k = 0 when k mod 3 ! = 0, and otherwise it is normal

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mean for the right hand side

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0great, i feel like i solved it. but you take credit for starting it

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but we didnt simplify it all the way

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0LHS we have 2^n / 3 + `

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh right, youre real part should be negative

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02pi/3 angle is in quadrant 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also x^n = 1 gives you the pentagonal shapes

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0or z^n = 1 , i should say

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0starting from (1,0) angle 0 , going counterclockwise

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ahhh.. memory so fickle art thou

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0quandrant 1 is 0 < theta < 90 q2 90 < theta < 180 , etc

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0q3 180 < theta < 270 , et

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh every nice, the 11 cancels the inside of the binomial

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02^n / 3 + ( i * sqrt 3 / 2 )^n + (i sqrt 3 / 2 ) ^n = n C0 + n C 3 + ..

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0notice that when you have n mod 3 = 0, then i^n + (i)^n both go to zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i^n + (i)^n is interesting

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it has the sequence for n = 1,2,3, 0 , 2, 0 , 2, 0, 2 , ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh thats because i^n +( i) ^n = i^n ( 1 + (1)^n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i mesed up , i didnt see the 1 + (1/2 + i sqrt 3 / 2 ) . so its

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0we have 2^n/3 + 1/3* ( 1 i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.02^n*1/3 + 1/3* ( 1 i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we can use an identity ( a + bi ) ^n + ( a  bi)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0where z* is the conjugate of z

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i hope im not boring you

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i might need some help here

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i did the case for i^n + (i)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sure...and what came up

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i^n ( 1  (1)^n ) , well a sequence

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for n = 1 ,2,3 we have 0,2,0,2,0,2 ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0lets try the case a + bi ^n + ( a bi)^n

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0yes that's what we're supposed to get remember with i the gap between terms is 2

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh the (a + bi)^n + ( a  bi)^n seems to diverge

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i bet you by looking at the binomial form, we can simplify , by looking at the series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so its easier if we divide out the a ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so ( 1 + ki ) ^n + ( 1  ki)^n ,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 + ki + (ki)^2 + k^3 *i^3 + ... 1 ki + (ki^2)  k^3i^3 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0= 1 + (ki)^2 + (ki)^4 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and we know 1 + x^2 + x^4 + ... = sum x^(2n)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so we get sum (ki)^(2n) for n=0...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0that is equal to 1 / ( 1 (ki)^2 )

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0( 1 + ki ) ^n + ( 1  ki)^n , = 1 / ( 1  (ki)^2 ) ?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0nevermind, this wont help us , lol

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0also i messed up, it should be ( 1 + ki ) ^n + ( 1  ki)^n = 2 + 2(ki)^2 + 2 (ki)^4 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh crap, i left out the binomial coefficients

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.01 + nC1*ki + nC2(ki)^2 +nC3 k^3 *i^3 + ... 1  nC0* ki + nC2* (ki^2) nC3 k^3i^3 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0actually the power series of complex variables shows some diverging oscillation type things

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0oh its an alternating series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0(( 1 + ki ) ^n + ( 1  ki)^n) /2 = 1  nC2 k^2 + nC4 k^4  nC6 k^6 + ...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i test it on k = 3, n = 4

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i dont know if this is .... useful formula. can we simplify the right side?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0might be a well known series

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so what were you saying earlier, how can we simplify this result, the conjugates w, and w^2 appear

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the final answer is then [ 2^n + ( 1/2  i sqrt (3 )/ 2 ) ^n + ( 1/2+ i sqrt (3) / 2 ) ^n ] / 3

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but if we can find a nice simplification of ( a + bi)^n + ( a  bi)^n, that would be nice

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0and this argument works for any prime p , because ( 1 + w + w^2 + ... w^p1) = 0

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0sorry but i dont see where the application is for this type of problem,

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0but its very pretty nonetheless
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