## Tuesday, 11 December 2012

### CBSE CLASS XI - MATHEMATICS - CH 8 - BINOMIAL THEOREM

BINOMIAL THEOREM

EXERCISE 8.1

Q.1 :  Expand the expression (1– 2x)5
ANS: By using Binomial Theorem, the expression (1– 2x)can be expanded as

Q.2 : Expand the expression
ANS :  By using Binomial Theorem, the expression  can be expanded as

Q.3 : Expand the expression (2x – 3)6
ANS : By using Binomial Theorem, the expression (2x – 3)can be expanded as

Q.4 : Expand the expression
ANS: By using Binomial Theorem, the expression  can be expanded as

Q.5 : Expand
ANS : By using Binomial Theorem, the expression  can be expanded as

Q.6 : Using Binomial Theorem, evaluate (96)3
ANS : 96 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, binomial theorem can be applied.
It can be written that, 96 = 100 – 4

Q.7 : Using Binomial Theorem, evaluate (102)5
ANS : 102 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 102 = 100 + 2

Q.8 : Using Binomial Theorem, evaluate (101)4
ANS : 101 can be expressed as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 101 = 100 + 1

Q.9: Using Binomial Theorem, evaluate (99)5
ANS : 99 can be written as the sum or difference of two numbers whose powers are easier to calculate and then, Binomial Theorem can be applied.
It can be written that, 99 = 100 – 1

Q.10: Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.
ANS : By splitting 1.1 and then applying Binomial Theorem, the first few terms of (1.1)10000 can be obtained as

Q.11 : Find (a + b)4 – (a – b)4. Hence, evaluate.
ANS : Using Binomial Theorem, the expressions, (a + b)4 and (a  b)4, can be expanded as

Q.12 : Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate.
ANS : Using Binomial Theorem, the expressions, (x + 1)6 and (x – 1)6, can be expanded as
By putting, we obtain

Q.13 : Show that is divisible by 64, whenever n is a positive integer.
ANS : In order to show that is divisible by 64, it has to be proved that,
, where k is some natural number
By Binomial Theorem,
For a = 8 and m = n + 1, we obtain
Thus, is divisible by 64, whenever n is a positive integer.

Q.14: Prove that.
ANS : By Binomial Theorem,
By putting b = 3 and a = 1 in the above equation, we obtain
Hence, proved.

EXERCISE 8.2

Q.1 : Find the coefficient of x5 in (x + 3)8

ANS: It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain
Comparing the indices of x in x5 and in Tr +1, we obtain
r = 3
Thus, the coefficient of x5 is

Q.2 : Find the coefficient of a5b7 in (a – 2b)12
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain
Comparing the indices of a and b in a5 band in Tr +1, we obtain
r = 7
Thus, the coefficient of a5b7 is

Q.3 : Write the general term in the expansion of (x2 – y)6
ANS : It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (+ b)n is given by .
Thus, the general term in the expansion of (x2 – y6) is

Q.4 : Write the general term in the expansion of (x2 – yx)12x ≠ 0
ANS : It is known that the general term Tr+1 {which is the (+ 1)th term} in the binomial expansion of (b)n is given by .
Thus, the general term in the expansion of(x2 – yx)12 is

Q.5 : Find the 4th term in the expansion of (x – 2y)12 .
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Thus, the 4th term in the expansion of (x – 2y)12 is

Q.6: Find the 13th term in the expansion of.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Thus, 13th term in the expansion of is

Q.7: Find the middle terms in the expansions of
ANS : It is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely, term and term.
Therefore, the middle terms in the expansion of are term and term
Thus, the middle terms in the expansion of are .

Q.8: Find the middle terms in the expansions of
ANS: It is known that in the expansion (a + b)n, if n is even, then the middle term is term.
Therefore, the middle term in the expansion of is term
Thus, the middle term in the expansion of is 61236 x5y5.

Q.9: In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that am occurs in the (r + 1)th term of the expansion (1 + a)m + n, we obtain
Comparing the indices of a in am and in T+ 1, we obtain
r = m
Therefore, the coefficient of am is
Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)m+n, we obtain
Comparing the indices of a in an and in Tk + 1, we obtain
k = n
Therefore, the coefficient of an is
Thus, from (1) and (2), it can be observed that the coefficients of am and an in the expansion of (1 + a)m + n are equal.

Q.10: The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of
(x + 1)n are in the ratio 1:3:5. Find n and r.
ANS : It is known that (+ 1)th term, (Tk+1), in the binomial expansion of (b)n is given by .
Therefore, (r – 1)th term in the expansion of (x + 1)n is
r th term in the expansion of (x + 1)n is
(r + 1)th term in the expansion of (x + 1)n is
Therefore, the coefficients of the (r – 1)thrth, and (r + 1)th terms in the expansion of (x + 1)n are  respectively. Since these coefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 and subtracting it from (2), we obtain
4– 12 = 0
⇒ r = 3
Putting the value of r in (1), we obtain
n – 12 + 5 = 0
⇒ n = 7
Thus, = 7 and r = 3

Q.11 : Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .

ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain
Comparing the indices of x in xn and in Tr + 1, we obtain
r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2– 1, we obtain
Comparing the indices of x in xn and Tk + 1, we obtain
k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2–1 is
From (1) and (2), it is observed that
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.

Q.12 : Find a positive value of m for which the coefficient of x2 in the expansion
(1 + x)m is 6.

ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (b)n is given by .
Assuming that x2 occurs in the (+ 1)th term of the expansion (1 +x)m, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Therefore, the coefficient of x2 is.
It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
Thus, the positive value of m, for which the coefficient of x2 in the expansion
(1 + x)m is 6, is 4.

Q.13: Show that is divisible by 64, whenever n is a positive integer.
ANS : In order to show that is divisible by 64, it has to be proved that,
, where k is some natural number
By Binomial Theorem,
For a = 8 and m = n + 1, we obtain
Thus, is divisible by 64, whenever n is a positive integer.

MISCELLANEOUS

Q.1 : Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
The first three terms of the expansion are given as 729, 7290, and 30375 respectively.
Therefore, we obtain
Dividing (2) by (1), we obtain
Dividing (3) by (2), we obtain
From (4) and (5), we obtain
Substituting n = 6 in equation (1), we obtain
a6 = 729
From (5), we obtain
Thus, a = 3, b = 5, and n = 6.

Q.2 : Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
ANS : It is known that (+ 1)th term, (Tr+1), in the binomial expansion of (+ b)n is given by .
Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Thus, the coefficient of x2 is
Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain
Comparing the indices of x in x3 and in Tk+ 1, we obtain
= 3
Thus, the coefficient of x3 is
It is given that the coefficients of x2 and x3 are the same.
Thus, the required value of a is.

Q.3 : Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
ANS : Using Binomial Theorem, the expressions, (1 + 2x)6 and (1 – x)7, can be expanded as
The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve x5, are required.
The terms containing x5 are
Thus, the coefficient of x5 in the given product is 171.

Q.4 : If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.
[Hint: write an = (a – b b)n and expand]
ANS : In order to prove that (a – b) is a factor of (an – bn), it has to be proved that
an – bn = k (a – b), where k is some natural number
It can be written that, a = a – b + b

This shows that (a – b) is a factor of (an – bn), where n is a positive integer.

Q.5 : Evaluate.
ANS : Firstly, the expression (a + b)6 – (a – b)6 is simplified by using Binomial Theorem.
This can be done as

Q.6 : Find the value of.
ANS : Firstly, the expression (x + y)4 + (x – y)4 is simplified by using Binomial Theorem.
This can be done as

Q.7: Find an approximation of (0.99)5 using the first three terms of its expansion.
ANS : 0.99 = 1 – 0.01

Thus, the value of (0.99)5 is approximately 0.951.

Q.8: Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
ANS : In the expansion, ,
Fifth term from the beginning
Fifth term from the end
Therefore, it is evident that in the expansion of, the fifth term from the beginning is and the fifth term from the end is.

It is given that the ratio of the fifth term from the beginning to the fifth term from the end is. Therefore, from (1) and (2), we obtain

Thus, the value of n is 10.

Q.9: Expand using Binomial Theorem.
ANS : Using Binomial Theorem, the given expression  can be expanded as

Again by using Binomial Theorem, we obtain

From (1), (2), and (3), we obtain

Q.10: Find the expansion of using binomial theorem.
ANS : Using Binomial Theorem, the given expression  can be expanded as

Again by using Binomial Theorem, we obtain

From (1) and (2), we obtain