Binomial Theorem and Important Terms of Binomial Expansion | Mathematics (Maths) Class 11 - Commerce PDF Download

Binomial Theorem

A. Binomial Theorem

The formula by which any positive integral power of a binomial expression can be expanded in the form of a series is known as Binomial Theorem. If  then

(x + y)ⁿ = ⁿC₀xⁿ + ⁿC₁x^{n – 1}y + ⁿC₂x^{n – 2} y² + ........+ ⁿC_rx^n–ry^r + ......+ ⁿC_nyⁿ 

This theorem can be proved by induction.

Observations : 

(a) The number of terms in the expansion is (n + 1) i.e. one more than the index.

(b) The sum of the indices of x & y in each term is n.

(c) The binomial coefficients of the terms (ⁿC₀, ⁿC₁.........) equidistant from the beginning and the end are equal.

Ex.1 The value of 

Sol. The numerator is of the form a³ + b³ + 3ab (a + b) = (a + b)³ where a = 18, and b = 7

 Nr = (18 + 7)³ = (25)³. Denominator can be written as

= (3 + 2)⁶ = 5⁶ = (25)³ 

B. IMPORTANT TERMS IN THE BINOMIAL EXPANSION ARE 

(a) General term : The general term or the (r + 1)^th term in the expansion of (x + y)ⁿ is given by

T_r+1 = ⁿC_rx^n-r. y^r 

Ex.2 Find : (a) The coefficient of x⁷ in the expansion of

(b) The coefficient of x^-7 in the expansion of

Also, find the relation between a and b, so that these coefficients are equal.

Sol. 

 (a) In the expansion of     

the general terms is 

putting 22 – 3r = 7 ⇒  3r = 15 ⇒  r = 5

Hence the coefficient of

(b) In the expansion of      ,

general terms is   

putting 11 – 3r = – 7 ⇒ 3r = 18  ⇒  r = 6

Hence the coefficient of

Also given coefficient of       

= coefficient of   

⇒  ab = 1 ( ∴ ¹¹C₅ = ¹¹C₆). Which is a required relation between a and b.

Hence the coefficient of x^–7 in  

Also given coefficient of x⁷ in 

¹¹C₅ a⁶ b^-5 = ¹¹C₆ a⁵b^-6 ⇒  ab = 1 (∵¹¹C₅ = ¹¹C₆). Which is a required relation between a and b.

 Ex.3 Find the number of rational terms in the expansion of (9^1/4 + 8^1/6)¹⁰⁰⁰. 

Sol. 

The general term in the expansion of (9^1/4 + 8^1/6)¹⁰⁰⁰ is T_{r + 1} = 

The above term will be rational if exponent of 3 and 2 are integers. i.e 1000-r/2 and r/2 must be integers

The possible set of values of r is {0, 2, 4, ..........1000}. Hence, number of rational terms is 501

(b) Middle term : The middle term(s) in the expansion of (x + y)ⁿ is (are)

(i) If n is even, there is only one middle term which is given by T_{(n + 2)/2} = ⁿC_n/2. x^n/2. y^n/2

(ii) If n is odd, there are two middle terms which are  &

Ex.4 Find the middle term in the expansion of 

Sol. 

The number of terms in the expansion of   is 10 (even). So there are two middle terms. 

 two middle terms. They are given by T₅ and T₆

(c) TERM INDEPENDENT OF x : Term independent of x contains no x ; Hence find the value of r for which the exponent of x is zero.

Ex.5 The term independent of x in 

Sol. General term in the expansion is 

For constant term, 3r/2 = 10, r = 20/3 which is not an integer. Therefore, there will be no constant term.

(d) NUMERICALLY GREATEST TERM : To find the greatest term in the expansion of (x + a)ⁿ.

We have   therefore, since xn multiplies every term in    , it will be sufficient to find the greatest term in this later expansion. Let the T_r and T_{r + 1} be the r^th and (r +1)^th terms in the expansion of     Let numerically, T_{r + 1} be the greatest term in the above expansion. Then

Substituting values of n and x, we get r < m + f or r < m where m is a positive integer and f is fraction such that 0 < f < 1. In the first case T_{m + 1} is the greatest term, while in the second case T_m and T_{m + 1} are the greatest terms and both are equal.

Ex.6 Find numerically the greatest term in the expansion of (3 – 5x)¹¹ when x = 1/5

Sol. Since 

so, the greatest terms are T_{2 + 1} and T₃₊₁.  Greatest term (when r = 2)

and greatest term (when r = 3)    

From above we say that the value of both greatest terms are equal.

C. If ( √A +B)n = I + f , where I & n are positive integers, n being odd and 0 < f < 1, then (I + f). f = Kⁿ where A – B² = K > 0 & A – B <1 . If n is an even integer, then (I + f) (1 – f) = kⁿ

Ex.7 If (6√6 + 14)²ⁿ⁺¹ = [N] + F and F = N – [N] ; where [*] denotes greatest integer, then NF is equal to

Sol. Since (6√6 + 14)²ⁿ⁺¹ = [N] + F. Let us assume that f = (6√6 – 14)²ⁿ⁺¹ ; where 0 ≤ f < 1.

[N] + F – f = even integer.

Now 0 < F < 1 and 0 < f < 1 so –1 < F – f < 1 and F – f is an integer so it can only be zero

Thus NF = (6√6 + 14)²ⁿ⁺¹ (6√6 – 14)²ⁿ⁺¹ = 20^{2n  + 1.}