Binomial Theorem, Chapter Notes, Class 11, Maths (IIT) PDF Download

binomial theorem

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 x, y Î R and n Î N, 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 is

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 T_{r + 1} =

putting 22 _ 3r = 7 Þ 3r = 15 Þ r = 5 T₆ =

Hence the coefficient of x⁷ in is ¹¹C₅ a⁶b^_5.

(b) In the expansion of , general terms is T_{r + 1} =

putting 11 _ 3r = _ 7 Þ 3r = 18 Þ r = 6

Hence the coefficient of x^_7 in is ¹¹C₆a⁵b^_6

Also given coefficient of x⁷ in = coefficient of x^_7 in

Þ Þ 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. and 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.

i.e. th and th two middle terms. They are given by T₅ and T₆

T₅ =T_{4 + 1} =

and

(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 is

Sol. General term in the expansion is

For constant term, Þ r= 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 (x + a)ⁿ = xⁿ ; 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 then . Let numerically, T_{r + 1} be the greatest term in the above expansion. Then T_{r + 1} ³ T_r Þ Þ Þ

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 (3 _ 5x)¹¹ = 3¹¹ . Now in the expansion of ,

we have =

Þ Þ 4r £ 12 Þ r £ 3 r = 2, 3

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

Ex.7 If = [N] + F and F = N _ [N] ; where [*] denotes greatest integer, then NF is equal to

Sol. Since = [N] + F. Let us assume that f = ; where 0 £ f < 1.

Now, [N] + F _ f = _

Þ [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 = = 20^{2n + 1}.

D. Some Results on Binomial Coefficients

(a) C₀ + C₁ + C₂ + ............+ C_n = 2ⁿ

(b) C₀ + C₂ + C₄ + ............= C₁ + C₃ + C₅ + .......... = 2^{n _ 1}

(c) C₀² + C₁² + C₂² + ............+ C_n² = ²ⁿC_n =

(d) C₀. C_r + C₁. C_{r + 1} + C₂. C_{r + 2} +............+C_{n _ r} C_n =

Remember : (2n) ! = 2ⁿ. n! [1.3.5........(2n _ 1)]

Ex.8 If (1 + x)ⁿ = C₀ + C₁x + C₂x² +.................+ C_nxⁿ then show that the sum of the products of the C_i¢s taken two at a time represents by : is equal to

Sol. Since (C₀ + C₁ + C₂ + ........+C_{n _ 1} + C_n)² .......+

Þ . Hence

Ex.9 If (1 + x)ⁿ = C₀ + C₁x + C₂x² + .........+C_nxⁿ then prove that

Sol. L.H.S = (C₀ + C₁)² + (C₀ + C₂)² + ....+ (C₀ + C_n)² + (C₁ + C₂)² + (C₁ + C₃)² + ....+ (C₁ + C_n)²

+ (C₂ + C₃)² + (C₂ + C₄)² + ........+ (C₂ + C_n)² + .........+ (C_{n_1} + C_n)²

= n (C₀² + C₁² + C₂² +..........+C_n²) + {from Ex. 8}

= n. ²ⁿC_n + 2²ⁿ _ ²ⁿC_n = (n _ 1). ²ⁿC_n + 2²ⁿ = R.H.S.

E. Binomial theorem for negative or fractional indices

If n Î Q, then (1 + x)ⁿ = 1 + nx + + +........... ¥ provided |x| < 1.

Note :

(i) When the index n is a positive integer the number of terms in the expansion of (1 + x)ⁿ is finite i.e. (n + 1) & the coefficient of successive terms are : ⁿC₀, ⁿC₁, ⁿC₂, .........., ⁿC_n

(ii) When the index is other than a positive integer such as negative integer or fraction, the number of terms in the expansion of (1 + x)ⁿ is infinite and the symbol ⁿC_r cannot be used to denote the coefficient of the general term.

(iii) Following expansion should be remembered (|x| < 1)

(a) (1 + x)^_1 = 1 _ x + x² _ x³ + x⁴ _ .........¥

(b) (1 _ x)^_1 = 1 + x + x² + x³ + x⁴ + .........¥

(c) (1 + x)^_2 = 1 _2x + 3x² _ 4x³ + .........¥

(d) (1 _ x)^_2 = 1 + 2x + 3x² + 4x³ + .........¥

(iv) The expansions in ascending powers of x are only valid if x is 'small'. If x is large i.e. |x| > 1 then we may find it convenient to expand in powers of 1/x, which then will be small.

F. Approximations

If x < 1, the terms of the above expansion go on decreasing and if x be very small, a stage may be reached when we may neglect the terms containing higher powers of x in the expansion. Thus, if x be so small that its squares and higher powers may be neglected then (1 + x)ⁿ = 1 + nx, approximately,

This is an approximate value of (1 + x)ⁿ

Ex.10 If x is so small such that its square and higher powers may be neglected then find the approximate value of

Sol.

= 1 _ _ = 1 _

Ex.11 The value of cube root of 1001 upto five decimal places is

Sol.

= 10 {1 + 0.0003333 _ 0.00000011 + ......} = 10.00333

Ex.12 The sum of is

Sol. Comparing with 1 + nx + + ...... Þ nx = 1/4 ...(i)

& or Þ Þ Þ ...(ii) {by (i)}

putting the value of x in (i) Þ Þ n =

sum of series = (1 + x)ⁿ = (1 _ 1/2)^_1/2 = (1/2)^_1/2 =

G. Exponential series

(a) e is an irrational number lying between 2.7 & 2.8. Its value correct upto 10 places of decimal is 2.7182818284.

(b) Logarithms to the base 'e' are known as the Napierian system, so named after Napier, their inventor. They are also called Natural Logarithm.

(c) where x may be any real or complex number &

(d) where a > 0

(e)

H. Logarithmic series

(a) ln (1 + x) = x _ where _ 1 < x £ 1

(b) ln (1 _ x) = where _ 1 £ x < 1

Remember : (i) (ii) e^{ln x} = x (iii) ln2 = 0.693 (iv) ln 10 = 2.303