Скачать или смотреть Coefficient of t^24 in the expansion of 〖(1+t^2)〗^12(1+t^12) (1+t^24) is 12C6 +2 b) 12C6 +1

   • Binomial Theorem Class 11  
Coefficient of t^24 in the expansion of 〖(1+t^2)〗^12(1+t^12) (1+t^24) is
12C6 +2 b) 12C6 +1 c) 12C6 d) none


In binomial theorem class 11, chapter 8 provides the information regarding the introduction and basic definitions for binomial theorem in a detailed way. To score good marks in binomial theorem class 11 concepts, go through the given problems here. Solve all class 11 Maths chapter 8 problems in the book by referring the examples to clear your concepts on binomial theorem.

Binomial Theorem Class 11 Topics
The topics and sub-topics covered in binomial theorem class 11 are:

Introduction
Binomial theorem for positive integral indices
Binomial theorem for any positive integer n
Special Cases
General and Middle Term
Binomial Theorem Class 11 Notes
Properties of Binomial Theorem for Positive Integer

(i) Total number of terms in the expansion of (x + a)n is (n + 1).
(ii) The sum of the indices of x and a in each term is n.

(iii) The above expansion is also true when x and a are complex numbers.

(iv) The coefficient of terms equidistant from the beginning and the end are equal. These coefficients are known as the binomial coefficients and

nCr = nCn – r, r = 0,1,2,…,n.

(v) General term in the expansion of (x + c)n is given by

Tr + 1 = nCrxn – r ar.

Binomial Expression: A binomial expression is an algebraic expression which contains two dissimilar terms. Ex: a + b, a3 + b3, etc.

Binomial Theorem: Let n ∈ N,x,y,∈ R then

nΣr=0 nCr xn – r · yr + nCr xn – r · yr + …………. + nCn-1 x · yn – 1 + nCn · yn

i.e.(x + y)n = nΣr=0 nCr xn – r · yr
some other useful expansions:

(x + y)n + (x−y)n = 2[C0 xn + C2 xn-1 y2 + C4 xn-4 y4 + …]
(x + y)n – (x−y)n = 2[C1 xn-1 y + C3 xn-3 y3 + C5 xn-5 y5 + …]
(1 + x)n = nΣr-0 nCr . xr = [C0 + C1 x + C2 x2 + … Cn xn]
(1+x)n + (1 − x)n = 2[C0 + C2 x2+C4 x4 + …]
(1+x)n − (1−x)n = 2[C1 x + C3 x3 + C5 x5 + …]
The number of terms in the expansion of (x + a)n + (x−a)n are (n+2)/2 if “n” is even or (n+1)/2 if “n” is odd.
The number of terms in the expansion of (x + a)n − (x−a)n are (n/2) if “n” is even or (n+1)/2 if “n” is odd.
Properties of Binomial Coefficients
Binomial coefficients refer to the integers which are coefficients in the binomial theorem. Some of the most important properties of binomial coefficients are:

C0 + C1 + C2 + … + Cn = 2n
C0 + C2 + C4 + … = C1 + C3 + C5 + … = 2n-1
C0 – C1 + C2 – C3 + … +(−1)n . nCn = 0
nC1 + 2.nC2 + 3.nC3 + … + n.nCn = n.2n-1
C1 − 2C2 + 3C3 − 4C4 + … +(−1)n-1 Cn = 0 for n greater than 1
C02 + C12 + C22 + …Cn2 = [(2n)!/ (n!)2]
General Term in binomial expansion:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn

General Term = Tr+1 = nCr xn-r . yr

General Term in (1 + x)n is nCr xr
In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .
Illustration: Find the number of terms in (1 + 2x +x2)50

Sol:

(1 + 2x + x2)50 = [(1 + x)2]50 = (1 + x)100

The number of terms = (100 + 1) = 101

Illustration: Find the fourth term from the end in the expansion of (2x – 1/x2)10

Sol:

Required term =T10 – 4 + 2 = T8 = 10C7 (2x)3 (−1/x2)7 = −960x-11

Middle Term(S) in the expansion of (x+y) n.n
If n is even then (n/2 + 1) Term is the middle Term.
If n is odd then [(n+1)/2]th and [(n+3)/2)th terms are the middle terms.
Illustration: Find the middle term of (1 −3x + 3x2 – x3)2n

Sol:

(1 − 3x + 3x2 – x3)2n = [(1 − x)3]2n = (1 − x)6n

Middle Term = [(6n/2) + 1] term = 6nC3n (−x)3n

Determining a Particular Term:

In the expansion of(axp + b/xq)n the coefficient of xm is the coefficient of Tr+1 where r = [(np−m)/(p+q)]
In the expansion of (x + a)n, Tr+1/Tr = (n – r + 1)/r . a/x
Independent Term
The term Independent of in the expansion of [axp + (b/xq)]n is

Tr+1 = nCr an-r br, where r = (np/p+q) (integer)

Illustration: Find the independent term of x in (x+1/x)6

Sol:

r = [6(1)/1+1] = 3

The independent term is 6C3 = 20
Numerically greatest term in the expansion of (1+x)n:
If [(n+1)|x|]/[|x|+1] = P, is a positive integer then Pth term and (P+1)th terms are numerically greatest terms in the expansion of (1+x)n
If[(n+1)|x|]/[|x|+1] = P + F, where P is a positive integer and 0 less F less 1 then (P+1)th term is numerically greatest term in the expansion of (1+x)n.