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Binomial Theorem is one of the most important chapters of Algebra in the syllabus of IIT JEE. The beginners sometimes find it difficult as this topic is very new to them. The other reason is that the concepts of Combinations are also used in this theorem. This chapter can be said to be one of the easiest as it has very few twists and turns.
The chapter begins with the Introduction to Binomial Theorem which is followed by the Properties of the Binomial Theorem. Binomial Coefficients are the most important topic of Binomial theorem. The sum of the Binomial coefficients, the coefficient of a particular term, greatest Binomial coefficient etc. has been discussed at length. However, some important results have been given at the end which is supplemented by the application of Binomial Expression. Solved examples as usual are very useful as they can reappear in the examination with slight modification and they also facilitate learning.
View the following video for more on Binomial Theorem
What is a Binomial Expression?
An algebraic expression consisting of two different terms is called a binomial expression. Eg: Terms of the form x^{3} + y^{3} are binomial expressions.
What is the Binomial Theorem for a positive integral?
The binomial theorem explains the way of expressing and evaluating the powers of a binomial. This theorem explains that a term of the form (a+b)^{n} can be expanded and expressed in the form of ra^{s}b^{t}, where the exponents s and t are non-negative integers satisfying the condition s + t = n. The coefficient r is a positive integer. The terms involved are called binomial coefficients and since it is for positive indices it expands only the positive powers.
The general binomial expansion for any index is given by
(x+y)^{n} = ^{n}C_{0}x^{n}y^{0} + ^{n}C_{1}x^{(n-1)}y^{1} + ^{n}C_{2}x^{(n-2)}y^{2 }+ …….. + ^{n}C_{(n-1)} x^{1}y^{(n-1)} + ^{n}C_{n}x^{0}y^{n}.
Introduction To Binomial Theorem
Properties Of Binomial Expansion
Binomial Coefficients
Sum Of Binomial Coefficients
Coefficient Of A Particular Term
Middle Term
Greatest Binomial Coefficient
Application Of Binomial Expression
Solved Examples
We shall be discussing these topics here in brief as they have been discussed in detail in the coming sections.
Number of terms in the expansion of (x + y)^{n} is (n + 1) i.e. one more than the index.
Sum of indices of x and y in each term in the expansion of (x + y)^{n} is n.
The binomial coefficients of the terms which are equidistant from the starting and the end are always equal. The simple reason behind this is
C (n, r) = C (n, n-r) which gives C (n, n) C (n, 1) = C (n, n-1) C (n, 2) = C (n, n-2).
We have ^{n}C_{r}x^{n-r}y^{r}
In this, if we replace y by -y, we have
^{n}C_{r}(-1)^{r} x^{n-r}. y^{r}
^{n}C_{r} x^{r}
^{ }(-1)^{r} ^{n}C_{r} x^{r}
(1 + x)^{n} + (1 - x)^{n} = 2[^{n}C_{0} + ^{n}C_{2}x^{2} + ….. ]
(1 + x)^{n} - (1 – x)^{n} = 2[^{n}C_{1} + ^{n}C_{3}x^{3} + ^{n}C_{5}x^{5} + …. ….. ]
Such an expansion always follows a simple rule which is
The subscript of C i.e. the lower suffix of C is always equal to the index of y and
Index of x = n – (lower suffix of C).
General Term
Term independent of x
Middle term
(a) General Term: T_{(r+1)}^{th} term is called as general term in (x + y)^{n} and the general term is given by
T_{(r+1)} = ^{n}C_{r}x^{n-r}y^{r}
(b) Term independent of x: It means the term containing x^{0}.
(c) Middle Term: Let T_{m} be the middle term in expansion of (x+y)^{n}, then
Case 1: If n is odd, then the number of terms will be even and so there will be two middle terms given by [(n+1)/2] th and [(n+3)/2] th.
Case 2: If n is even, then the number of terms will be odd and so there will be only one middle term i.e. (n/2 + 1)th.
Note: The binomial coefficient of the middle term is the greatest binomial coefficient of the expansion.
Numerically greatest Term in (x+y)^{n}: T_{r+1} term is said to be numerically greatest for a given value of x, y provided T_{r+1} ≥ T_{r} and T_{r+1} ≥ T_{r+2}
i.e.
T_{r+1}/T_{r} = ^{n}C_{r}x^{n-r}y^{r}/^{n}C_{r-1}(x)^{n-r+1}y^{r-1} = (n-r+1)/r |(y/x)|
Properties of ^{n}C_{r}:
^{n}C_{r} = ^{n}C_{n-r} menas ^{n}C_{x} = ^{n}C_{y} has two solutions x = y or x + y = n.
^{n}C_{r} + ^{n}C_{r-1} = ^{n+1}C_{r}
^{n}C_{r} = n/r (^{n-1}C_{r-1})
^{n}C_{r}/^{n}C_{r-1} = (n-r+1)/r
In order to find the integral part of the above expression, follow the below listed steps:
Step 1: Consider N’ = (a-√b)n or (√b-a)n according as a > √b or √b > a.
Step 2: Use N + N’ or N – N’ such that the result is an integer.
Step 3: Use the fact N = I + f, where I stands for [N] and ‘f’ for {N}.
Let us have a look at some of questions that can be asked from this section:
Illustration 1: (a) Find the coefficient of x^{201} in the expansion of 1 + (1 + x) + (1 + x)^{2} + ….. (1 + x)^{2011}.
(b) Find the coefficient of x^{301} in the expansion of (1 + x) + 2(1 + x)^{2} + 3(1 + x)^{3} + ….. … + 500 (1 + x)^{500}.
Solution: 1 + (1 + x) + (1 + x)^{2} + ….. (1 + x)^{2011}
So, the coefficient of x^{201} in 1/x . (1 + x)^{2012} – 1/x
⇒ The coefficient of x^{202} in (1 + x)^{2012} ⇒ ^{2012}C_{202}.
(b) S = (1 + x) + 2(1 + x)^{2} + 3(1 + x)^{3} + ….. … + 500 (1 + x)^{500}. ….….. (1)
(1 + x)S = (1 + x)^{2} + 2(1 + x)^{3} + ….. … + 499 (1 + x)^{500} + 500 (1 + x)^{501} ….….. .(2)
Subtracting (2) from (1), we have
S(-x) = (1 + x) + (1 + x)^{2} + (1 + x)^{3} + ….. … + (1 + x)^{500} – 500(1 + x)^{501}
= (1 + x) [{(1 + x)^{500} – 1}/ (1 + x – 1)] – 500(1 + x)^{501}
S(-x) = (1 + x)[(1 + x)^{500} – 1]/ x – 500(1 + x)^{501}
So, S = 1/x^{2} (1 + x)(1 – (1 + x)^{500}) + 500. 1/x . (1 + x)^{501}
= 1/x^{2 }- (1 + x)^{500}/x^{2} + 1/x – (1 + x)^{500}/x + 500. 1/x (1 + x)^{501}
So, coefficient of x^{301} = – ^{500}C_{303} – ^{500}C_{302} + 500. ^{501}C_{302}
= 500. ^{501}C_{302} – (^{500}C_{303} + ^{500}C_{302})
= 500. ^{501}C_{302} – ^{501}C_{303}.
Illustration 2: In the binomial expansion of (a-b)^{n}, n ≥ 5, the sum of the 5th and 6th terms is zero. Then find the value of a/b.
Solution: The sum of the 5th term is given by
T_{5} = ^{n}C_{4}a^{n-4}(-b)^{4}
The sum of the 6th term is given by
T_{6} = ^{n}C_{5}a^{n-5}(-b)^{5}
It is given in the question that T_{5} + T_{6} = 0.
This gives a/b = (n-4)/5.
Illustration 3: Prove that C_{0} – 2^{2}C_{1} + 3^{2}C_{2} - ….. + (-1)^{n}(n+1)^{2} C_{n} = 0, n > 2, where C_{r} = ^{n}C_{r}.
Solution: We know that by the binomial theorem
(1 + x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + …. + C_{n}x^{n}
Multiplying it by x we get,
x(1 + x)^{n} = C_{0}x + C_{1}x^{2} + C_{2}x^{3} + ……. + C_{n}x^{n+1}
Differentiating both sides we get,
(1 + x)^{n} + nx(1 + x)^{n-1} = C_{0} + 2C_{1}x + 3C_{2}x^{2} + ….. + (n+1)C_{n}x
Again multiplying by x we get,
x(1+x)^{n} + nx^{2}(1+x)^{n-1} = C_{0}x + 2C_{1}x^{2} + 3C_{2}x^{3} + ……. + (n+1)C_{n}x^{n+1}
Hence, (1 + x)^{n} + nx(1 + x)^{n-1} + 2 nx(1 + x)^{n-1} + n(n-1)x^{2}(1 + x)^{n-2} = C_{0} + 2^{2}C_{1}x + 3^{2}C_{2}x^{2} + ….. + (n + 1)^{2} C_{n}x_{n}
Putting x = -1, we get
0 = C_{0} – 2^{2}C_{1} + 3^{2}C_{2} - ….. + (-1)^{n}(n+1)^{2}C_{n}
Illustration 4: If (1+x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + C_{3}x^{3} + …. + C_{n}x^{n}, then prove that C_{1}^{2} + 2C_{2}^{2} + 3C_{3}^{2} + ….….…... + nC_{n}^{2 }= (2n-1)!/[(n-1)!]^{2}
Solution: Given (1+x)^{n} = C_{0} + C_{1}x + C_{2}x^{2} + C_{3}x^{3} + …. + C_{n}x^{n}
Differentiating both sides with respect to x, we get
n(1+x)^{n-1} = 0 + C_{1} + 2C_{2}x + 3C_{3}x^{2} + …. + nC_{n}x^{n-1}
n(1+x)^{n-1} = C_{1} + 2C_{2}x + 3C_{3}x^{2} + …. + nC_{n}x^{n-1} ….… (1)
and (x+1)^{n} = C_{0}x^{n} + C_{1}x^{n-1} + C_{2}x^{n-2 }+ C_{1}x^{n-1 }+ ….… + C_{n} ….….(2)
Multiplying (1) and (2), we have
n(1 + x)^{2n-1} = (C_{1} + 2C_{2}x + 3C_{3}x^{2} + …. + nC_{n}x^{n-1}) x (C_{0}x^{n} + C_{1}x^{n-1} + C_{2}x^{n-2} + C_{3}x^{n-3} + ….. + C_{n}) ….. (3)
Now, coefficient of x^{n-1} on R.H.S.
C_{1}^{2} + 2C_{2}^{2} + 3C_{3}^{2} + ….. + nC_{n}^{2}
and coefficient of x^{n-1} on L.H.S = n.^{2n-1}C_{n-1}
But, we know that (3) is an identity and therefore the coefficient of x^{n-1} in R.H.S. = coefficient of x^{n-1} in L.H.S.
So, C_{1}^{2} + 2C_{2}^{2} + 3C_{3}^{2} + …. + nC_{n}^{2} = (2n-1)!/{(n-1)!}.
Q1. The number of terms in the expansion of (x + y)^{n} is
(a) one more than the index
(b) one less than the index
(c) same as index
(d) none of these
Q2. ^{n}C_{r} + ^{n}C_{r-1} =
(a) ^{n}Cr-1
(b) ^{n+1}C_{r}
(c) ^{n+1}C_{r+1}
(d) ^{n+1}C_{r-1}
Q3. The general term in (x + y)^{n} is given by
(a) T_{r+1} = ^{n}C_{r}x^{n-r.}y^{r}
(b) T_{r+1} = ^{n}C_{r-1}x^{n-r.}y^{r}
(c) T_{r+1} = ^{n}C_{r}x^{n-r.}y^{r}
Q4. The middle term in expansion of (x + y)^{n}, in case of n being odd is
(a) (n+3)/2 th
(b) n/2 th
(c) (n/2 + 1) th
(d) (n+1)/2 th and (n+3)/2 th
Q5. The middle term in expansion of (x + y)^{n}, in case of n being even is
(a) (n/2 + 1) th
(c) (n+1)/2 th
Binomial Theorem is important from the perspective of scoring high in IIT JEE as there are few fixed pattern on which a number Multiple Choice Questions are framed on this topic. The chapters of Binomial Theorem, Permutations and Combinations, and Probability together fetch 10-20 marks varying from one examination to the other.
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