Question

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

Answer 1

it goes on until nCn

Answer 2

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

Answer 3

p.s. what courses are you studying?

Answer 4

ive just graduated from year 12

Answer 5

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

Answer 6

i think ill get into one of the iits

Answer 7

no problem

Answer 8

so r u indian?

Answer 9

wt are you doing??

Answer 10

i dont see an answer

Answer 11

hes right...that druvidfae is bloody bright

Answer 12

z^3 = 1 , how does that help ?

Answer 13

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

Answer 14

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

Answer 15

get it??

Answer 16

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

Answer 17

but roots of unity are not real

Answer 18

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

Answer 19

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

Answer 20

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

Answer 21

sorry had to leave for a moment

Answer 22

hey druvidfae wtr u studying>?

Answer 23

but n is not given?

Answer 24

the answer is in terms of n

Answer 25

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.

Answer 26

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)

Answer 27

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

Answer 28

where w^3 = 1

Answer 29

= 0

Answer 30

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

Answer 31

woops

Answer 32

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)

Answer 33

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

Answer 34

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

Answer 35

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

Answer 36

whats the left side of this expression?

Answer 37

do you have an equation ?

Answer 38

i get 3 + 0 + 0 + ...

Answer 39

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.

Answer 40

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

Answer 41

whoa whoa , slow down there buddy

Answer 42

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

Answer 43

you guys leave out too many steps.

Answer 44

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

Answer 45

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

Answer 46

right

Answer 47

yes

Answer 48

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

Answer 49

( 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))

Answer 50

IS THAT WHAT YOU ARE DOING ????

Answer 51

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

Answer 52

the left side is a closed formula?

Answer 53

i would like you guys to actually solve it, lol

Answer 54

im still learning this stuff, sorry

Answer 55

i messed up

Answer 56

( 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))

Answer 57

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.

Answer 58

ok , im a little worried about the right side though

Answer 59

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

Answer 60

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

Answer 61

( 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))

Answer 62

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

Answer 63

although n is not necessarily a multiple of 3

Answer 64

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

Answer 65

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

Answer 66

also we have to assume that n | 3

Answer 67

errr, 3 | n

Answer 68

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

Answer 69

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

Answer 70

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

Answer 71

oh you mean before dividing by 3 ?

Answer 72

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.

Answer 73

we have deletion of terms

Answer 74

oh right

Answer 75

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

Answer 76

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

Answer 77

e^ ( i * pi/3 ) ?

Answer 78

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

Answer 79

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

Answer 80

yes your forms are also correct

Answer 81

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

Answer 82

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

Answer 83

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

Answer 84

oh its easier if we start with theta = 0

Answer 85

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

Answer 86

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

Answer 87

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

Answer 88

so w and w^2 are conjugates, interesting

Answer 89

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

Answer 90

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

Answer 91

now test, let n = 8

Answer 92

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

Answer 93

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 ?

Answer 94

or last term should be 8 C 6

Answer 95

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

Answer 96

yes

Answer 97

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

Answer 98

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