Class 11 — Mathematics (NCERT)

Binomial Theorem

Class 11

✓By the end of this lesson students will be able to understand the Binomial Theorem for positive integral indices.
✓By the end of this lesson students will be able to expand binomial expressions using the theorem.
✓By the end of this lesson students will be able to determine the general term in a binomial expansion.
✓By the end of this lesson students will be able to find the middle term(s) in a binomial expansion.
✓By the end of this lesson students will be able to apply the Binomial Theorem to solve related problems.

Key concepts

Binomial Expression

An algebraic expression containing two terms is called a binomial expression. For example, (x+y), (2x-3y), (x^2 + 1/x) are binomial expressions.

Binomial Theorem for Positive Integral Indices

The Binomial Theorem provides a formula for expanding any positive integral power of a binomial expression. For any positive integer 'n', the expansion of (a+b)^n is given by:\n(a+b)^n = nC0 a^n b^0 + nC1 a^(n-1) b^1 + nC2 a^(n-2) b^2 + ... + nCr a^(n-r) b^r + ... + nCn a^0 b^n\nThis can be written using summation notation as:\n(a+b)^n = Σ (from r=0 to n) nCr a^(n-r) b^r\nwhere nCr (read as 'n choose r') is the binomial coefficient, defined as:\nnCr = n! / (r! * (n-r)!)\nand n! (n factorial) = n * (n-1) * (n-2) * ... * 2 * 1. Also, 0! = 1.\n\nProperties of Binomial Expansion (a+b)^n:\n1. The total number of terms in the expansion is (n+1).\n2. The sum of the indices of 'a' and 'b' in each term is always 'n'.\n3. The binomial coefficients nC0, nC1, ..., nCn are symmetric, i.e., nCr = nC(n-r).\n4. The coefficients are the numbers in Pascal's Triangle.

(a+b)^n = Σ (from r=0 to n) nCr a^(n-r) b^r

General Term (T_r+1)

The (r+1)th term in the binomial expansion of (a+b)^n is called the general term and is denoted by T_r+1. This formula is very useful for finding any specific term in the expansion without writing out the entire expansion.

T_r+1 = nCr a^(n-r) b^r

Middle Term(s)

The position of the middle term(s) depends on whether 'n' (the power) is even or odd.\n1. If n is even: There is only one middle term. Its position is the (n/2 + 1)th term. So, the middle term is T_(n/2 + 1).\n2. If n is odd: There are two middle terms. Their positions are the ((n+1)/2)th term and the ((n+3)/2)th term. So, the middle terms are T_((n+1)/2) and T_((n+3)/2).

Key facts to remember

1The Binomial Theorem for a positive integer n is (a+b)^n = Σ (from r=0 to n) nCr a^(n-r) b^r.
2The binomial coefficient nCr is given by nCr = n! / (r! * (n-r)!).
3The total number of terms in the expansion of (a+b)^n is (n+1).
4The general term (or (r+1)th term) in the expansion of (a+b)^n is T_r+1 = nCr a^(n-r) b^r.
5If n is even, there is one middle term, which is the (n/2 + 1)th term.
6If n is odd, there are two middle terms, which are the ((n+1)/2)th and ((n+3)/2)th terms.
7Important properties of binomial coefficients: nC0 = 1, nCn = 1, nC1 = n, and nCr = nC(n-r).

Worked examples

Example 1

Expand (2x - 3y)^4.

IHere, a = 2x, b = -3y, and n = 4.

IIUsing the Binomial Theorem: (a+b)^n = nC0 a^n + nC1 a^(n-1) b + nC2 a^(n-2) b^2 + nC3 a^(n-3) b^3 + nC4 a^0 b^4

IIISubstitute the values:

IV(2x - 3y)^4 = 4C0 (2x)^4 (-3y)^0 + 4C1 (2x)^3 (-3y)^1 + 4C2 (2x)^2 (-3y)^2 + 4C3 (2x)^1 (-3y)^3 + 4C4 (2x)^0 (-3y)^4

VCalculate binomial coefficients: 4C0 = 1, 4C1 = 4, 4C2 = 6, 4C3 = 4, 4C4 = 1.

VISubstitute and simplify each term:

VII= 1 * (16x^4) * 1

VIII+ 4 * (8x^3) * (-3y)

9+ 6 * (4x^2) * (9y^2)

10+ 4 * (2x) * (-27y^3)

11+ 1 * 1 * (81y^4)

12Perform multiplication:

13= 16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

Answer

16x^4 - 96x^3y + 216x^2y^2 - 216xy^3 + 81y^4

Pay close attention to the signs of the terms, especially when 'b' is negative.

Example 2

Find the coefficient of x^7 in the expansion of (3x^2 - 1/(2x))^11.

IHere, a = 3x^2, b = -1/(2x), and n = 11.

IIThe general term T_r+1 is given by: T_r+1 = nCr a^(n-r) b^r

IIISubstitute the values:

IVT_r+1 = 11Cr (3x^2)^(11-r) (-1/(2x))^r

VSeparate the constant and variable parts:

VIT_r+1 = 11Cr * 3^(11-r) * (x^2)^(11-r) * (-1/2)^r * (1/x)^r

VIIT_r+1 = 11Cr * 3^(11-r) * (-1/2)^r * x^(22-2r) * x^(-r)

VIIIT_r+1 = 11Cr * 3^(11-r) * (-1/2)^r * x^(22-3r)

9For the coefficient of x^7, we equate the power of x to 7:

1022 - 3r = 7

113r = 15

12r = 5

13Substitute r = 5 into the general term to find the coefficient:

14Coefficient = 11C5 * 3^(11-5) * (-1/2)^5

15Coefficient = 11C5 * 3^6 * (-1/32)

16Calculate 11C5 = 11! / (5! * 6!) = (11 * 10 * 9 * 8 * 7) / (5 * 4 * 3 * 2 * 1) = 462.

17Calculate 3^6 = 729.

18Substitute and simplify:

19Coefficient = 462 * 729 * (-1/32)

20Coefficient = - (336798) / 32

21Coefficient = - 168399 / 16

Answer

The coefficient of x^7 is -168399/16.

Always simplify the powers of x (or the variable) first to find the correct 'r' value.

Example 3

Find the middle term(s) in the expansion of (x/3 + 9y)^10 and (x^2 + 1/x)^7.

I**Part 1: Expansion of (x/3 + 9y)^10**

IIHere, n = 10, which is an even number.

IIISince n is even, there is only one middle term. Its position is (n/2 + 1)th term.

IVPosition = (10/2 + 1) = (5 + 1) = 6th term.

VSo, we need to find T_6. Using the general term formula T_r+1 = nCr a^(n-r) b^r, we have r+1 = 6, so r = 5.

VIHere, a = x/3, b = 9y, n = 10, r = 5.

VIIT_6 = 10C5 (x/3)^(10-5) (9y)^5

VIIIT_6 = 10C5 (x/3)^5 (9y)^5

9Calculate 10C5 = 10! / (5! * 5!) = (10 * 9 * 8 * 7 * 6) / (5 * 4 * 3 * 2 * 1) = 252.

10Substitute and simplify:

11T_6 = 252 * (x^5 / 3^5) * (9^5 * y^5)

12T_6 = 252 * (x^5 / 243) * (59049 * y^5)

13T_6 = 252 * x^5 * y^5 * (59049 / 243)

14T_6 = 252 * x^5 * y^5 * 243

15T_6 = 61236 x^5 y^5

16**Part 2: Expansion of (x^2 + 1/x)^7**

17Here, n = 7, which is an odd number.

18Since n is odd, there are two middle terms. Their positions are ((n+1)/2)th term and ((n+3)/2)th term.

19Position 1 = ((7+1)/2) = 4th term.

20Position 2 = ((7+3)/2) = 5th term.

21For the 4th term (T_4): r+1 = 4, so r = 3.

22a = x^2, b = 1/x, n = 7, r = 3.

23T_4 = 7C3 (x^2)^(7-3) (1/x)^3

24T_4 = 7C3 (x^2)^4 (1/x)^3

25T_4 = 7C3 x^8 x^(-3)

26T_4 = 7C3 x^5

277C3 = 7! / (3! * 4!) = (7 * 6 * 5) / (3 * 2 * 1) = 35.

28T_4 = 35x^5

29For the 5th term (T_5): r+1 = 5, so r = 4.

30a = x^2, b = 1/x, n = 7, r = 4.

31T_5 = 7C4 (x^2)^(7-4) (1/x)^4

32T_5 = 7C4 (x^2)^3 (1/x)^4

33T_5 = 7C4 x^6 x^(-4)

34T_5 = 7C4 x^2

357C4 = 7! / (4! * 3!) = (7 * 6 * 5) / (3 * 2 * 1) = 35 (since 7C4 = 7C3).

36T_5 = 35x^2

Answer

For (x/3 + 9y)^10, the middle term is 61236 x^5 y^5. For (x^2 + 1/x)^7, the middle terms are 35x^5 and 35x^2.

Remember to check if 'n' is even or odd to correctly determine the number and position of middle terms.

Common mistakes

✗Incorrectly handling negative signs in 'a' or 'b' terms, leading to sign errors in the expansion.
✗Errors in calculating binomial coefficients (nCr), especially for larger values of n.
✗Mistakes in applying exponent rules, particularly with terms involving fractions or negative powers (e.g., (x^m)^n, 1/x^k).
✗Confusing the term number (k) with the 'r' value in the general term formula T_r+1 (i.e., for the k-th term, r = k-1).
✗Incorrectly identifying the position of middle terms, especially when 'n' is an odd number and two middle terms are required.

Exam tips

★Memorise the general term formula (T_r+1 = nCr a^(n-r) b^r) as it is crucial for solving most problems related to specific terms.
★Always clearly identify 'a', 'b', and 'n' from the given binomial expression. Pay close attention to any negative signs.
★Practice calculating nCr values quickly and accurately to save time during the examination.
★When finding specific terms (e.g., term independent of x, coefficient of x^k), always set up the general term first and then equate the power of the variable to the required value to find 'r'.
★For middle terms, first determine if 'n' is even or odd to know if there's one or two middle terms, then correctly find their positions before calculating the terms.

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