In the expansion of (x^2+1+(1)/(x^2))^n, n ∈𝐍, (a) number of terms is 2 n+1 (b) coefficient of c

In the expansion of (x^2+1+(1)/(x^2))^n, n ∈𝐍, (a) number of...

Key Concepts

Multinomial Theorem

The multinomial theorem generalizes the binomial theorem to sums with three or more terms. It provides a systematic way to expand expressions like (a + b + c)^n by assigning multinomial coefficients that represent the number of ways to choose the powers of each term such that the exponents add up to n. This theorem is fundamental for understanding how to create and organize terms in the expansion of any multinomial expression.

Symmetry in Polynomial Expansions

Symmetry in polynomial expansions arises when the terms in the expression exhibit a balanced structure, such as having exponents that are opposites of one another. This symmetry can facilitate the combination of like terms and can often simplify the process of counting distinct terms or determining specific coefficients. Recognizing symmetry is an essential skill when handling expansions where opposing exponents or mirror-image terms are present.

Exponent Combination Techniques

In polynomial expansions involving terms with variable exponents, determining the exponent of a specific term in the expansion relies on combining the exponents contributed by each factor. This involves setting up equations that relate the counts of how many times each term is chosen to achieve a desired overall exponent. Mastering these techniques allows one to map out all possible exponent outcomes in the expansion and understand the structure of the resulting polynomial.

Counting Distinct Terms

Counting the distinct terms in an expansion requires an understanding of how the choices from each summand combine and whether some combinations yield like terms. In expressions where the exponents vary by specific increments, one can determine the number of distinct powers present by examining the range and step size. This concept is crucial for assessing not just the total number of terms before combining similar ones, but also the number of unique exponents after simplification.

Coefficient Determination in Multinomial Expansions

Coefficient determination in multinomial expansions involves calculating the sum of all multinomial coefficients associated with terms that contribute to a particular power. This process typically includes identifying all combinations of selections that yield the same overall exponent and summing their corresponding coefficients. Accurate coefficient determination is essential for fully describing the terms in the expansion and for solving problems that ask for the numerical factors of specific terms.

Transcript

00:02 Hi, i'm just given here this extension here the term where an analog natural numbers we're going to find out here a number of terms the coefficient of the constant term coefficient of extra part 2 and negative 2 and the coefficient of x squared so first and to expand this series here so we can rewrite this as this was given as 1 plus x square plus 1 over x square to the power and this will take a single term here expand it it's coming out to be we get nc 0 plus nc1 x squared plus 1 over x squared plus nc2 x square plus 1 over x squared whole square and so on going till nc n x squared plus 1 over x squared to the park this is the first term here 1 to any power barely one only so they're not consider to consider that so that's the one only now let's take here the terms here we have so here the term is extra power 0 term is extra power 2 extra power 4 and so on the terms here which contain here that is given as extra power 0 extra power 2 extra 4 and so on now it's going till and term so we get here extra part 2m so here we have extra part 2n the more terms that we have here that is going to be extra negative 2 here extra by negative 4 extra per negative 2nd 2nd so that is given as extra negative 2 x2 x2 by negative 4 and so on we have extra negative 2 s this we have all terms here if we now find out here the first part of the question is the coefficient of the constant term so just working on this here the coefficient of constant term if you look at that so where we are the constant term so here we'll have a constant term because we have a square plus b square plus 2 a b square plus 2 a b so x square and x square will cancel out we'll have a constant term here here we have a constant term so we have the coefficient of constant term or coefficient of constant term that is given as nc 0 next we have nc2 nc2 and 2 mean this term here and c2 and from here get 2 a b so we get 2 here nc 2 times next term will be nc 4 and look at here nc4 after that we have nc4 which will make constant term if i look at here the nc4 terms so i will get it as in nc4 it will be x square plus 1 over x square to the power 4 so we got nc4 after this if we just extend this so the third term we will have the constant term here so with expand this here third term will look at a third term that will be 4 c2 we have x square to the part 2 then 1 over x square with a part 2 you can see here that cancels out x2 4 x2 and we get this 4 c2 here let's also multiply here so we get here times 4 c2 next will be nc 6 and it will be 6 3 so on so clearly this look at this series here that is not equal to two to the part n negative to one next next just we have finite a coefficient of x2 2 and negative 2 so for that if you observe here the last term that contains extra part 2 n so for 2 and negative 2 n must be n negative 1 that means we're talking about here nc n negative 1 term x2 plus 1 over x squared with the power n negative 1 and extend this for the first term for this the first term would be it will be n negative 1 p 0 x squared of the power we have n negative 1 and then 1 over x2 to the power 0 so you can see here that's community 1 here you get x2 2 and negative 2 that is required x2 and negative 2 and this is common to be 1.

05:11 So coefficient is coming out to be nc n negative 1.

05:15 And if you divide this, that is coming out to be nc r equals nc and negative 1, that will be nc1, that is equal to.

05:27 The next part we have the coefficient of x square.

05:29 So for that, we just see this here, we get x -quire coefficient from here, that is nc1.

05:39 Next we get the coefficient of x square if you look at for nc3 term.

05:44 So, if i look at here, nc3 term here we have, nc3 times x squared plus 1 over x squared to the whole cube.

05:52 So we just write as nc3 times, if i take here a second term from those series, so 3c1, x squared to the power 3 negative 1, 1 over x squared to the power 1.

06:02 Now simplify this, i will get again x squared.

06:04 So i get this coefficient...

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