# Question-1

**Expand each of the following expression (1 â€“ 2x)**

^{5}**Solution:**

By using Binomial Theorem, we have

(1 â€“2x)

^{5}= [1+(-2x)]

^{5}

=

^{5}C

_{0}+

^{5}C

_{1}(-2x) +

^{5}C

_{2}(-2x)

^{2}+

^{5}C

_{3}(-2x)

^{3}+

^{5}C

_{4}(-2x)

^{4}+

^{5}C

_{5}(-2x)

^{5}

= 1 + 5 (-2x) + 10 (-2x )

^{2}+ 10 (-2x)

^{3 }+ 5(-2x)

^{4}+ (-2x)

^{5 }

= 1 â€“ 10x + 40x

^{2}- 80x

^{3}+ 80x

^{4}- 32x

^{5}

# Question-2

**Expand each of the following expression**

**Solution:**

By using Binomial Theorem, we have

=

= ^{5}C_{0}^{5}C_{1} + ^{5}C_{2+ }^{5}C_{3 + }^{5}C_{4+ }^{5}C_{5 }

=

=

# Question-3

**Expand each of the following expression (2x â€“ 3)**

^{6 }**Solution:**

By using Binomial Theorem, we have

(2x -3)

^{6}= [2x +(-3)]

^{6}

=

^{6}C

_{0}(2x)

^{6}(-3)Â° +

^{6}C

_{1}(2x)

^{5}(-3)

^{1}+

^{6}C

_{2}(2x)

^{4}(-3)

^{2}+

^{6}C

_{3}(2x)

^{3}(-3)

^{3}+

^{6}C

_{4}(2x)

^{2}(-3)

^{4}+

^{6}C

_{5}(2x)

^{1}(-3)

^{5}+

^{6}C

_{6}(2x)

^{0}(-3)

^{6}

= 1(2x)

^{6}+ 6(2x)

^{5}(-3) + 15(2x)

^{4}(-3)

^{2}+ 20(2x)

^{3}(-3)

^{3}+ 15(2x)

^{2}(-3)

^{4}+ 6(2x)(-3)

^{5}+(-3)

^{6}

= 64x

^{6}â€“ 576x

^{5}+ 2160x

^{4}â€“ 4320x

^{3}+ 4860x

^{2}â€“ 2916x + 729

# Question-4

**Expand each of the following expression**

**Solution:**

By using Binomial Theorem, we have;

=

^{5}C

_{0}+

^{5}C

_{1}+

^{5}C

_{2}+

^{5}C

_{3}+

^{5}C

_{4}+

^{5}C

_{5}

=

=

# Question-5

**Expand each of the following expression**

**Solution:**

By using Binomial Theorem, we have:

=

^{5}C

_{0}(x)

^{6}+

^{6}C

_{1}(x)

^{5}+

^{6}C

_{2}(x)

^{4}+

^{6}C

_{3}(x)

^{3}+

^{6}C

_{4}(x)

^{2}+

^{6}C

_{5}(x)

^{1}+

^{6}C

_{6}

** = **x^{6} + 6x^{5
}l_{ }= x^{6} + 6x^{4} + 15x^{2} + 20 +

# Question-6

**Using binomial theorem, evaluate each of the following (96)**

^{3}**Solution:**

We express 96 as the sum or difference of two numbers whose powers are easier to calculate, and then use Binomial Theorem

Write 96 = 100 - 4

Therefore

(96)

^{3}= (100 - 4)

^{3}

=

^{3}C

_{0}(100)

^{3 }-

^{3}C

_{1}(100)

^{2}(4) +

^{3}C

_{2}(100)

^{1}(4)

^{2 }-

^{3}C

_{3}(4)

^{3 }

= 1000000 -3 (10000) (4) + 3 (100) (16) - (64)

= 1000000 - 120000 + 4800 - 64

= 884736

# Question-7

**Using binomial theorem, evaluate each of the following (102)**

^{5}**Solution:**

(102)

^{5}= (100 + 2)

^{5 }

= ^{5}C_{0}(100)^{5 }+ ^{5}C_{1}(100)^{4}(2) + ^{5}C_{2}(100)^{3}(2)^{2 }+ ^{5}C_{3}(100)^{2}(2)^{3 }+ ^{5}C_{4}(100)(2)^{4 }+ ^{5}C_{5}(2)^{5}

=10000000000 + 5(100000000)(2)+10(1000000)(4)+ 10(10000)(8) + 5(100)(16) + 32

= 10000000000+1000000000 + 40000000 + 800000 + 8000 + 32

= 11040808032

# Question-8

**Using binomial theorem, evaluate each of the following (101)**

^{4}**Solution:**

(101)

^{4}= (100+ 1)

^{4 }

= ^{4}C_{0}(100)^{4} + ^{4}C_{1}(100)^{3}(1)^{ }+ ^{4}C_{2}(100)^{2}(1)^{2} +^{4}C_{3}(100)^{1}(1)^{3} +^{4}C_{4}(1)^{4}

= 100000000 + 4(1000000) + 6 (10000)+ 4 (100)+ 1

= 100000000 + 4000000 + 60000 + 400+1

= 104060401

# Question-9

**Using binomial theorem, evaluate each of the following (99)**

^{5}**Solution:**

(99)

^{5}= (100- 1)

^{5}

=

^{5}C

_{0}(100)

^{5 }-

^{5}C

_{1}(100)

^{4}.1 +

^{5}C

_{2}(100)

^{3}.(1)

^{2 }-

^{5}C

_{3}(100)

^{2}.(1)

^{2 }+

^{5}C

_{4}(100)

^{1}(1)

^{4 }-

^{5}C

_{5}(1)

^{5 }

= (100)

^{5 }â€“ 5 Ã— (100)

^{4 }+ 10 Ã—(100)

^{3 }â€“ 10 Ã— (100)

^{2 }+ 5 Ã— 100 â€“ 1

= 10000000000-500000000+ 10000000 - 100000 + 500 - 1

= 10010000500 - 500100001

= 9509900499

# Question-10

**Using Binomial Theorem indicate which number is larger (1.1)**

^{10000}or 1000.**Solution:**

Splitting 1.1 and using Binomial Theorem to write the first few terms we have

(1.1)

^{10000}= (1+0.1)

^{10000 }

=

^{10000}C

_{0}+

^{10000}C

_{1}(0.1) +

^{10000}C

_{2}(0.1)

^{2}+ other positive terms.

= 1 + 10000 Â´ (0.1) + other positive terms

= 1 + 1000 + other positive terms

>1000

Hence, (1.1)

^{10000 }> 1000.

# Question-11

**Find (a + b)**

^{4}- (a â€“ b)^{4}. Hence, evaluate .**Solution:**

(a + b)

^{4}-(a - b)

^{4}

= [

^{4}C

_{0}a

^{4}+

^{4}C

_{1}a

^{3}b +

^{4}C

_{2}a

^{2}b

^{2}+

^{4}C

_{3}ab

^{3}+

^{4}C

_{4}b

^{4}] - [

^{4}C

_{0}a

^{4 + 4}C

_{l}a

^{3}b +

^{4}C

_{2}a

^{2}b

^{2}-

^{4}C

_{3}ab

^{3}+

^{4}C

_{4}b

^{4}] = 2 Ã—

^{4}C

_{1}a

^{3}b + 2 Ã—

^{4}C

_{3}ab

^{3}

= 2[4a

^{3}b + 4ab

^{3}]

= 8ab [a

^{2}+ b

^{2}]

Thus, ( +)

^{4}- ( - )

^{4}

= 8 . [3 + 2] = 8(5) = 40

# Question-12

**Find (x + 1)**

^{6}+ (x - 1)^{6}. Hence or otherwise evaluate**Solution:**

(x + 1)

^{6}+ (x - 1)

^{6}

= [

^{6}C

_{0}x

^{6}+

^{6}C

_{1}x

^{5}+

^{6}C

_{2 }x

^{4}] + [

^{6}C

_{3}x

^{3}+

^{6}C

_{4}x

^{2}+

^{6}C

_{5}x

^{1}-

^{6}C

_{6}] +

[

^{6}C

_{0}x

^{6}-

^{6}C

_{1}x

^{5}+

^{6}C

_{2}x

^{4}-

^{6}C

_{3 }x

^{3}+

^{6}C

_{4}x

^{2 }+

^{6}C

_{5}x

^{1 }â€“

^{6}C

_{6}]

= 2[

^{6}C

_{0}x

^{6}+

^{6}C

_{2}x

^{4}+

^{6}C

_{4}x

^{2}+

^{6}C

_{6}]

= 2[x

^{6}+ 15x

^{4}+ 15x

^{2}+ 1]

Thus, (+ 1)

^{6}+ (- 1)

^{6}

= 2 [()

^{6}

**+**15()

^{4}

**+**15 () + 1]

= 2 [8+ 15 (4)+ 15 (2) + 1]

= 2 [8 + 60 + 30+ 1]= 198.

# Question-13

**Show that 9**

^{n+1}- 8n - 9 is divisible by 64, whenever n is a positive integer.**Solution:**

n = 1 â‡’ 9

^{n + 1}- 8n - 9 = 9

^{2}- 8 - 9

= 81 - 17 = 64= 1(64)

n = 2 â‡’9

^{n + 1}- 8n - 9 = 9

^{3}- 8(2) - 9

= 729 â€“ 16 - 9 = 704= 11 (64)

From n = 3, 4, 5,.....9

^{n + 1}â€“ 8n - 9 = 9(1 + 8)n - 8n - 9

= 9 [

^{n}C

_{0}+

^{n}C

_{1}. 8 +

^{n}C

_{2}.8

^{2}+ ...

^{n}C

_{n}8

^{n}] â€“ 8n - 9

= 9[1 + 8n +

^{n}C

_{2}.8

^{2}+ ...

^{n}C

_{n}8n] â€“8n â€“ 9

= 9 + 72n + 9.

^{n}C

_{2}.

_{ }8

^{2}+ ... 9

^{n}C

_{n}8

^{n}â€“8n - 9

= 8

^{2}[n + 9 (

^{n}C

_{2}+

^{n}C

_{3}.8 +...

^{n}C

_{n}8

^{n-2})]

which is divisible by 64.

# Question-14

**Prove that**3

^{r}

^{n}C

_{r}= 4

^{n }

_{.}

**Solution:**

L.H.S = 3

^{0}C(n,0) + 3

^{1}C(n,1) + 3

^{2}C(n,2) + --- 3

^{r}(n, r) + --- +3

^{n}C(n, n)

= C(n,0) + C(n,1) 3

^{1}+ C(n,2) 3

^{2}+ C(n,3) 3

^{3}+ --- +C(n,n)3

^{n}

This is in the form of (1+3)

^{n}

= (1+3)

^{n}= 4

^{n }= R.H.S

# Question-15

**Prove that x**

^{5}in (x + 3)^{8}**Solution:**

Suppose x

^{5}occurs in the (r + 1 )th term of the expansion (x + 3)

^{8 }

Now T

_{r+1}=

^{n}C

_{r}a

^{n - r}b

^{r}=

^{8}C

_{r }x

^{8 - r}3

^{r}

Comparing the indices of x in x

^{5}and in T

_{r + 1}, we get r = 3

Thus, the coefficient of x

^{5}is

^{8}C

_{3}(3)

^{3}= 1512

# Question-16

**Prove that a b**

^{7}in (a-2b)^{12}**Solution:**

Let a

^{5}b

^{7}occurs in the (r + 1)th term, in the expansion of (a - 2b)

^{12}given by

^{I2}C

_{r}. a

^{12 â€“ r}(-2b)

^{r}. Then 12 â€“ r = 5. This gives r= 7.

Thus the

**coefficient of**

**a**

^{5}b

^{7}is

^{12}C

_{5}(-2)

^{7}= (-128) = (792) (-128) = -101376

# Question-17

**Prove that (x**

^{2}â€“ y)^{6}**Solution:**

We have T

_{r + 1}in (a + b)

^{n}=

^{n}C

_{r}a

^{n - r}. b

^{r},0 â‰¤ r â‰¤ n

T

_{r + 1 }in (x

^{2 }- y)

^{6}=

^{6}C

_{r}(x

^{2})

^{6 - r }(-y)

^{r }

=

^{ 6}C

_{r }x

^{12 }â€“

^{2r}(-y)

^{r}

# Question-18

**Prove that (x**

^{2}â€“ yx)^{12}, x Â¹ 0**Solution:**

T

_{r + 1}in (x

^{2}-yx)

^{12}=

^{12}C

_{r}(x

^{2})

^{12 - r}(-yx)

^{r}

=

^{12}C

_{r}x

^{24 â€“ 2r }(-1)

^{r}(y)

^{r}(x)

^{r}

=

^{12}C

_{r}x

^{24 â€“ r }y

^{r}(-1)

^{r}

# Question-19

**Find the 4th term in the expansion of (x - 2y)**

^{12}.**Solution:**

4th term in (x - 2y)

^{12}= T

_{4}= T

_{3}

_{+ 1}

=

^{12}C

_{3}(x)

^{12 -}

^{3}(-2y)

^{3}

[T

_{r + 1}in (a + b)

^{n}=

^{n}C

_{r}a

^{n â€“ r}b

^{r}]

=

^{12}C

_{3}(x)

^{9 }(-2)

^{3}(y)

^{3 }

=

= -1760 x

^{9}y

^{3}

# Question-20

**Find the 13**^{th}term in the expansion of**Solution:**

13

^{th}term in = T

_{13 }= T

_{12 + 1}

=

^{18}C

_{12}(9x)

^{18 - 12}

=

^{18}C

_{12}(9x)

^{6}

= ^{18}C_{12}(9x)^{6}

= ^{18}C_{12}(9x)^{6}

= ^{18}C_{12}(3^{2})^{6}

= ^{18}C_{12}(3^{12})^{}

= ^{18}C_{12 }= 18564

# Question-21

**Find the 13**

^{th}term in the expansion of**Solution:**

The index of, is 7, which is an odd natural number.

So, Middle terms are and

= T_{4} = T_{3+1} = ^{7}C_{3}(3)^{7 â€“ 3 }

= ^{7}C_{3}(3)^{4}(x^{3})^{3}(-1)(6^{-1})^{3}

= ^{7}C_{3}(81)(x)^{9} (6)^{-3}(-1)^{3}

= =

= T_{5} = T_{4} + 1 = ^{7}C_{4}(3)^{7 â€“ 4 }

= ^{7}C_{4}(3)^{3 }(-1)^{4}(x^{3})(6)^{-4}

= ^{7}C_{4}(27)^{ }(x)^{12}(6)^{-4}

= (35)(27)(6)^{-4}(x)^{12}

= =

# Question-22

**Solution:**

The index is 10, which is an even natural number.

Hence, Middle term = = T_{6} = T_{5+1}

= ^{10}C_{5(9y)}^{5}

= ^{10}C_{5(9y)}^{5}

= ^{10}C_{5}(x)^{5}_{}

= ^{10}C_{5}(3)^{-5}_{}

= ^{10}C_{5}(3)^{-5}_{}

= ^{10}C_{5}3^{5 }x^{5 }y^{5} = (252)(243)x^{5} y^{5}

= 61236 x^{5}y^{5}

# Question-23

**In the expansions of (1 + a)**

^{m+n}, using Binomial Theorem, prove that coefficients of a^{m}and a^{n}are equal.**Solution:**

We have,

(1 + a)^{m+n} = [^{m+n}C_{0} + ^{m+n}C_{1a}^{1} + ^{m+n}C_{2a}^{2}+â€¦â€¦ ^{m+n}C_{r}

a^{r} + â€¦â€¦ + ^{m+n}C_{m+n} a^{m+n}

Coefficient of a^{m} =^{ m+n}C_{m} =

Also the coefficient of a^{n}

= ^{m+n}C_{n} =

Clearly, ^{m+n}C_{m} = ^{m+n}C_{n}

# Question-24

**The coefficients of the (r-1)**

^{th}, r^{th}and (r+1)^{th}terms in the expansion of (x+1)^{n}are in the ratio 1:3:5. Find both n and r.**Solution:**

Coefficient of (r-1)th term = C(n, r-2)

Coefficient of rth term = C(n, r-1)

Coefficient of (r+1)th term = C(n, r)

Considering 1

^{st}and

^{2nd}

3r - 3 = n - r + 2

n - 4r = -5 ----------(1)

Considering 2nd and 3rd

5r = 3n - 3r +3

3n - 8r = -3 ---------(2)

2(n - 4r = -5)

2n - 8r = -10 ---------(3)

Subtract (3) from (2)

n = 7

Substitute n = 7 in (2)

We get r = 3

n = 7, r = 3

# Question-25

**Prove that the coefficient of x**

^{n}in the expansion of (1+x)^{2n}is twice the coefficient of x^{n}in the expansion (1 + x)^{2n â€“ 1}.**Solution:**

(1+x)

^{2n}=

^{2n}C

_{0}+

^{2n}C

_{1}x

^{1}+

^{2n}C

_{2}x

^{2}+â€¦. +

^{2n}C

_{n}x

^{n}

(1+x)

^{2n-1}=

^{2n-1}C

_{0}+

^{2n-1}C

_{1}x

^{1}

_{ }+

^{2n-1}C

_{2}x

^{2}+â€¦. +

^{2n-1}C

_{n}x

^{n}

Coefficient of x^{n} in (1+x)^{2n â€“ 1} is (^{2n - 1}C_{n})

^{2n}C_{n} =

=

=

= 2 â€¦â€¦.(i)

Now ^{2n - 1}C_{n} =

=

=

=

= â€¦â€¦.(ii)

From (i) and (ii) we have

^{2n}C_{n} = 2. ^{2n-1}C_{n}