anonymous
  • anonymous
nc0 + nc3 + nc6 ......Please help
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
it goes on until nCn
anonymous
  • 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
  • anonymous
p.s. what courses are you studying?

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anonymous
  • anonymous
ive just graduated from year 12
anonymous
  • anonymous
neat.....which college/university/institute are you trying for
anonymous
  • anonymous
i think ill get into one of the iits
anonymous
  • anonymous
no problem
anonymous
  • anonymous
so r u indian?
anonymous
  • anonymous
wt are you doing??
anonymous
  • anonymous
i dont see an answer
anonymous
  • anonymous
hes right...that druvidfae is bloody bright
anonymous
  • anonymous
z^3 = 1 , how does that help ?
anonymous
  • anonymous
u put in the cube roots of unity one by one and add all three equations...the required series comes out
anonymous
  • anonymous
because all other binomial coefficients like nc1 and nc2 and so on have a (1 + w + w^2) multiplied
anonymous
  • anonymous
get it??
anonymous
  • anonymous
ok so ( 1 + z^1/3 ) ( 1 + z^1/3)( 1 + z ^1/3)
anonymous
  • anonymous
but roots of unity are not real
anonymous
  • anonymous
yes..first put in 1, then w and then w^2..u dont have to use only real nos ;)
anonymous
  • anonymous
one sec, im a bit rusty here, ( 1 + x ) ^m = 1 + mx + m(m-1)/2 x^2 + ..
anonymous
  • anonymous
same way if i wanted nC0 + nC5 + nC10... i would have put in the fifth roots of unity..
anonymous
  • anonymous
sorry had to leave for a moment
anonymous
  • anonymous
hey druvidfae wtr u studying>?
anonymous
  • anonymous
but n is not given?
anonymous
  • anonymous
the answer is in terms of n
anonymous
  • 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
  • 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
  • anonymous
oh wait, so youre using the fact that ( 1 + w + w^2 ) = ...
anonymous
  • anonymous
where w^3 = 1
anonymous
  • anonymous
= 0
anonymous
  • 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
  • anonymous
woops
anonymous
  • 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
  • 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
  • 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
  • anonymous
ok, and w^4 = w , that sort of thing
anonymous
  • anonymous
whats the left side of this expression?
anonymous
  • anonymous
do you have an equation ?
anonymous
  • anonymous
i get 3 + 0 + 0 + ...
anonymous
  • 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
  • anonymous
canto:yes... the principle can be extended to other problems as well.
anonymous
  • anonymous
whoa whoa , slow down there buddy
anonymous
  • anonymous
first thing is first, youre using the identity ( 1 + 1) ^n = Sum nCk , from k=0 to n ?
anonymous
  • anonymous
you guys leave out too many steps.
anonymous
  • anonymous
youre doing ( 1 + w)^n, (1+1)^n, and ( 1 + w^2) ^n, adding them
anonymous
  • anonymous
for this question we don't need the 2^n identity
anonymous
  • anonymous
right
anonymous
  • anonymous
yes
anonymous
  • anonymous
( 1 + w)^n + (1+1)^n, + ( 1 + w^2) ^n = 3 + n( 1 + w + w^2) + nC2 ( 1 + w^2 + w^4 ) + ...
anonymous
  • 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
  • anonymous
IS THAT WHAT YOU ARE DOING ????
anonymous
  • anonymous
and then stuff vanishes on the right side, but the left side i dont know what it is
anonymous
  • anonymous
the left side is a closed formula?
anonymous
  • anonymous
i would like you guys to actually solve it, lol
anonymous
  • anonymous
im still learning this stuff, sorry
anonymous
  • anonymous
i messed up
anonymous
  • 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
  • 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
  • anonymous
ok , im a little worried about the right side though
anonymous
  • anonymous
= 3 + 0 + 0 + nC3 + 0 + 0
anonymous
  • anonymous
why all non-3 get reduced to 0 . that is the sequence we want
anonymous
  • 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
  • anonymous
so the final answer is [2^n + (1+w)^n + (1+w^2)^n ]/3 = nC0 + n C 3 + ... n C n
anonymous
  • anonymous
although n is not necessarily a multiple of 3
anonymous
  • anonymous
why all non-3 get reduced to 0 . that is the sequence we want
anonymous
  • anonymous
im just checking , do you agree with my closed form formula ?
anonymous
  • anonymous
also we have to assume that n | 3
anonymous
  • anonymous
errr, 3 | n
anonymous
  • anonymous
yeah the l.h.s. reduces to something like 3^n (note there are probably some more factors i'm missing)
anonymous
  • anonymous
hmmm, no that doesnt make sense, it should be less than 2^n
anonymous
  • anonymous
since nC0 + nC1 + ... n Cn = 2^n
anonymous
  • anonymous
oh you mean before dividing by 3 ?
anonymous
  • 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
  • anonymous
we have deletion of terms
anonymous
  • anonymous
oh right
anonymous
  • anonymous
then you would zeroes as the last 1 or two terms ?
anonymous
  • anonymous
ok im going to test this , offhand i dont know the third root of unity of 1
anonymous
  • anonymous
e^ ( i * pi/3 ) ?
anonymous
  • 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
  • anonymous
1 = e^(2*Pi *i) , so 1^(1/3) = e^(2pi/3 *i)
anonymous
  • anonymous
yes your forms are also correct
anonymous
  • anonymous
sorry, i have to start from fundamental principles, i have horrible memory
anonymous
  • anonymous
1 = e^( i*[ 2*Pi + 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3 +2pi/3*n)
anonymous
  • anonymous
so 1^1/3 = e ^ ( i * (2pi/3 , 4pi/3
anonymous
  • anonymous
oh its easier if we start with theta = 0
anonymous
  • anonymous
1 = e^( i*[ 0+ 2pi *n ) , so 1^(1/3) = e^(i * [2pi/3*n)
anonymous
  • anonymous
so we have z^(1/3) = { 1 , e^(i*2pi/3) , e^ ( i*4pi/3)
anonymous
  • anonymous
i assume you have it memorized down well :) so i agree with that
anonymous
  • anonymous
so w and w^2 are conjugates, interesting
anonymous
  • anonymous
but it might be more succinct calculator wise to use exponential polar form
anonymous
  • anonymous
[2^n + (1+e^(i*2pi/3))^n + (1+e^(i*4pi/3))^n ]/3 = nC0 + n C 3 + ... n C n
anonymous
  • anonymous
now test, let n = 8
anonymous
  • anonymous
sorry my early values have a slight correction the real part should be negative.
anonymous
  • 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
  • anonymous
or last term should be 8 C 6
anonymous
  • anonymous
does your calculator allow complex calculations......i.e. calculations involving iota based calculations.
anonymous
  • anonymous
yes
anonymous
  • anonymous
i get 85 for left side 85 + 3.3666666666666666666666E -13i
anonymous
  • anonymous
the last term should be\[(n & 6)\]
anonymous
  • anonymous
yessssssssssss
anonymous
  • anonymous
it works
anonymous
  • anonymous
*nc6
anonymous
  • anonymous
if n = 8
anonymous
  • anonymous
8C0 + 8C 3 + 8 C 6 = 85
anonymous
  • anonymous
so the imaginary part vanishes there
anonymous
  • anonymous
haha thats funny, my calcualto has 3.33333333336 E -13 i
anonymous
  • anonymous
so that vanishes , its like 3.3 *10^-13 * i
anonymous
  • anonymous
for the left side its 85 + 3.3666666666666666666666 E -13i E means 10^
anonymous
  • anonymous
which calculator do you have ??
anonymous
  • anonymous
why didnt him actually solve it, thats annoying
anonymous
  • anonymous
TI 84
anonymous
  • anonymous
its easy to use, and fast to learn , the TI 89 is more powerful, but harder to use
anonymous
  • anonymous
im pissed, maple cant solve tan (105)
anonymous
  • anonymous
its an exact angle
anonymous
  • anonymous
TI 89 has multiple color codes, so light blue, light green for functions
anonymous
  • anonymous
my calculator:casio fx82 and fx 100 my numerical software :octave
anonymous
  • anonymous
it should be nC0 + nC3 + ... n C n divisible by 3
anonymous
  • anonymous
or something like n C k where k mod 3 is not equal to zero
anonymous
  • anonymous
not necessarily (due to the additional 2^n factor)
anonymous
  • anonymous
sum nC k , where nC k = 0 when k mod 3 ! = 0, and otherwise it is normal
anonymous
  • anonymous
i mean for the right hand side
anonymous
  • anonymous
yeah on the r.h.s.
anonymous
  • anonymous
great, i feel like i solved it. but you take credit for starting it
anonymous
  • anonymous
but we didnt simplify it all the way
anonymous
  • anonymous
LHS we have 2^n / 3 + `
anonymous
  • anonymous
oh right, youre real part should be negative
anonymous
  • anonymous
2pi/3 angle is in quadrant 2
anonymous
  • anonymous
also x^n = 1 gives you the pentagonal shapes
anonymous
  • anonymous
or z^n = 1 , i should say
anonymous
  • anonymous
northeast quadrant
anonymous
  • anonymous
northwest quadrant
anonymous
  • anonymous
starting from (1,0) angle 0 , going counterclockwise
anonymous
  • anonymous
ahhh.. memory so fickle art thou
anonymous
  • anonymous
quandrant 1 is 0 < theta < 90 q2 90 < theta < 180 , etc
anonymous
  • anonymous
q3 180 < theta < 270 , et
anonymous
  • anonymous
oh every nice, the 1-1 cancels the inside of the binomial
anonymous
  • anonymous
2^n / 3 + ( i * sqrt 3 / 2 )^n + (-i sqrt 3 / 2 ) ^n = n C0 + n C 3 + ..
anonymous
  • anonymous
notice that when you have n mod 3 = 0, then i^n + (-i)^n both go to zero
anonymous
  • anonymous
no i take that back
anonymous
  • anonymous
i^n + (-i)^n is interesting
anonymous
  • anonymous
it has the sequence for n = 1,2,3, 0 , -2, 0 , 2, 0, -2 , ...
anonymous
  • anonymous
oh thats because i^n +(- i) ^n = i^n ( 1 + (-1)^n)
anonymous
  • anonymous
i mesed up , i didnt see the 1 + (-1/2 + i sqrt 3 / 2 ) . so its
anonymous
  • anonymous
we have 2^n/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n
anonymous
  • anonymous
2^n*1/3 + 1/3* ( 1- i sqrt 3 / 2 ) ^n + ( 1 + i sqrt 3 / 2 ) ^n
anonymous
  • anonymous
and we can use an identity ( a + bi ) ^n + ( a - bi)^n
anonymous
  • anonymous
or z^n + z*^n = ...
anonymous
  • anonymous
where z* is the conjugate of z
anonymous
  • anonymous
i hope im not boring you
anonymous
  • anonymous
no definately not
anonymous
  • anonymous
i might need some help here
anonymous
  • anonymous
i did the case for i^n + (-i)^n
anonymous
  • anonymous
sure...and what came up
anonymous
  • anonymous
i^n ( 1 - (-1)^n ) , well a sequence
anonymous
  • anonymous
for n = 1 ,2,3 we have 0,-2,0,2,0,-2 ...
anonymous
  • anonymous
lets try the case a + bi ^n + ( a -bi)^n
anonymous
  • anonymous
yes that's what we're supposed to get remember with i the gap between terms is 2
anonymous
  • anonymous
oh the (a + bi)^n + ( a - bi)^n seems to diverge
anonymous
  • anonymous
i bet you by looking at the binomial form, we can simplify , by looking at the series
anonymous
  • anonymous
so its easier if we divide out the a ,
anonymous
  • anonymous
so ( 1 + ki ) ^n + ( 1 - ki)^n ,
anonymous
  • anonymous
oh very nice we get
anonymous
  • anonymous
1 + ki + (ki)^2 + k^3 *i^3 + ... 1 -ki + (ki^2) - k^3i^3 + ...
anonymous
  • anonymous
= 1 + (ki)^2 + (ki)^4 + ...
anonymous
  • anonymous
and we know 1 + x^2 + x^4 + ... = sum x^(2n)
anonymous
  • anonymous
so we get sum (ki)^(2n) for n=0...
anonymous
  • anonymous
that is equal to 1 / ( 1- (ki)^2 )
anonymous
  • anonymous
( 1 + ki ) ^n + ( 1 - ki)^n , = 1 / ( 1 - (ki)^2 ) ?
anonymous
  • anonymous
only if |ki|< 1
anonymous
  • anonymous
nevermind, this wont help us , lol
anonymous
  • anonymous
also i messed up, it should be ( 1 + ki ) ^n + ( 1 - ki)^n = 2 + 2(ki)^2 + 2 (ki)^4 + ...
anonymous
  • anonymous
oh crap, i left out the binomial coefficients
anonymous
  • anonymous
1 + nC1*ki + nC2(ki)^2 +nC3 k^3 *i^3 + ... 1 - nC0* ki + nC2* (ki^2) -nC3 k^3i^3 + ...
anonymous
  • anonymous
actually the power series of complex variables shows some diverging oscillation type things
anonymous
  • anonymous
oh its an alternating series
anonymous
  • anonymous
(( 1 + ki ) ^n + ( 1 - ki)^n) /2 = 1 - nC2 k^2 + nC4 k^4 - nC6 k^6 + ...
anonymous
  • anonymous
yay it works
anonymous
  • anonymous
i test it on k = 3, n = 4
anonymous
  • anonymous
i dont know if this is .... useful formula. can we simplify the right side?
anonymous
  • anonymous
might be a well known series
anonymous
  • anonymous
so what were you saying earlier, how can we simplify this result, the conjugates w, and w^2 appear
anonymous
  • anonymous
the final answer is then [ 2^n + ( 1/2 - i sqrt (3 )/ 2 ) ^n + ( 1/2+ i sqrt (3) / 2 ) ^n ] / 3
anonymous
  • anonymous
yes
anonymous
  • anonymous
but if we can find a nice simplification of ( a + bi)^n + ( a - bi)^n, that would be nice
anonymous
  • anonymous
and this argument works for any prime p , because ( 1 + w + w^2 + ... w^p-1) = 0
anonymous
  • anonymous
sorry but i dont see where the application is for this type of problem,
anonymous
  • anonymous
but its very pretty nonetheless

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