If n is a multiple of 3 and (1+x+x^2)^n=∑r=0^2 n ap xp and ∑r=0^n(-1)^r ar( ^n Cr)=k( ^n Cn 3) t

If n is a multiple of 3 and (1+x+x^2)^n=∑r=0^2 n ap xp and...

Key Concepts

Binomial Coefficient Identities

Binomial coefficients, as featured in the Binomial Theorem, possess many useful identities that simplify the manipulation of sums and expansions. These identities help in rewriting sums involving binomial terms into more manageable forms, and often reveal underlying combinatorial relationships.

Coefficient Extraction

Extracting the coefficient of a specific power from a polynomial expansion is a common technique in combinatorics. It involves identifying the term that corresponds to a certain degree, thereby translating an algebraic expression into a combinatorial count or identity.

Polynomial Multiplication and Convolution

When two generating functions (or polynomials) are multiplied, the coefficient of a given term in the product is given by a convolution of the original sequences. This process helps in transforming a complicated sum over products of coefficients into a single coefficient in a new series, which can sometimes be easier to evaluate.

Generating Functions

Generating functions are an algebraic tool used to encode sequences as power series. They allow us to manipulate infinite series to extract useful combinatorial information, such as finding relationships between coefficients or summing sequences encoded within a polynomial.

Transcript

00:01 Hi, it is given that here and is a multiple of three and we have one plus x plus x per to the bar n is given as the summation here summation are equal to n a xr and we're also given here the summation is given as equal to k and cn over 3 to find the value of k here what you can do we just start here with this expression here summation r equal 0 to n so write this as summation r equal 0 to 0 to n negative 1 the power r a r we have n z here now we'll observe this what the summation means here before that we'll look at two series here one series we'll look at that is given as a notch plus a1x plus a 2x squared and so on we have a 2n x2n x2 2n so this city is starting from here 0 1 2 until this is taken the series here.

01:11 Of course we're given here this 1 plus x squared to the power 1 plus x plus x2 to the power n this is given as summation r equal to 0 2 n ar xr so you use this so we get here it's going to be a naught plus a 1x plus a 2 x plus a 2 x and so on till a 2n x2n so this summation we've got here and expanded this series you got this series here now the next series so we'll take now that's as even as and the wording in the next series see here we require a r ncr so let's say here we have it is a 1x so we require something that is which multiply with this so it should be nc1 1 over x right so it's just multiply here we have let's say this a 1x take here multiply with we have nc 1 1 over x it just multiply this we get a 1 and c1 so we're getting this term here that means the next series which we will take and we multiply with this to get this summation here, that series will take it as, so we take it as nc0, negative nc1, 1 over x, positive nc2, 1 over x squared, and so on, we have negative 1 to the power n times nc n, 1 over x to the power n.

02:38 The series we're getting here.

02:41 Now we just multiply here this two series.

02:46 So this is let's say, as s1 here was given as s2 so it's just multiply here s1 times s2 and we just take here multiply then and we take here the coefficient of the constant term because we have we get here here negative a 1 nc1 there we get a 2 and c2 x squared with 1 over x2 so we just take here the coefficient of s 1 s2 then that will give us this series from r equal to 0 0 0 to n negative 1 to the power r, ar, ncr, ncr.

03:22 So we can say the series here, summation r equals 0 to n, negative 1 to the power r, ar, ncr, is actually the coefficient we have.

03:38 Coefficient of s1 times s2.

03:44 This series here, a 0 plus a1x and a2x square.

03:49 This given as we have summation r equal to 0 2n, and arxr.

03:54 So this is given as 1 plus x plus x squared to the power n.

04:00 So we get s1 that is coming out to be a little product here for s1 that is coming out to be 1 plus x plus x square to the power n and this series here and 0 negative and c1 1 over x if you place here x with 1 over x.

04:19 So in the series we have let's say 1 plus x to the power n if replace x with negative sign so we get alternate negative sign here and if we just replace x with 1 over x so we get this series so this is given as we get here the series s2 that is coming up to be times one negative 1 over x to the power n and now to simplify this we get 1 plus x plus x squared to the power n times x negative 1 to the power n so maybe here at negative 1 to the power n over x to par n now this series here this one is a coefficient of constant term in this series here so if i just take this extra par n onto this side so this will become the coefficient of x2 par n because x2 p .m is dividing here so it will multiply by the coefficient here so this series here required series will will become the coefficient of x to power n in the series x square plus x plus 1 to the bar n times x negative 1 to the power n.

05:46 You can see in a different way also...

Please give Ace some feedback