Class XI CBSE-Mathematics Binomial Theorem · 2019-11-27 · Class–XI–CBSE-Mathematics Binomial Theorem 5 243 + 5 3 81 + 10 27 + 10 9 + 5 3 3 1 5 5. Expand the expression ( +1

Class–XI–CBSE-Mathematics Binomial Theorem

CBSE NCERT Solutions for Class 11 Mathematics Chapter 08

Back of Chapter Questions

Exercise 8.1

1. Expand the expression (1 − 2𝑥)5

Hint: ∑ 𝐶𝑛𝑟

𝑛

𝑟=0𝑎𝑛−𝑟𝑏𝑟 = (𝑎 + 𝑏)𝑛

Solution:

Solution step 1: ((1 − 2𝑥)5

Expression (1 − 2𝑥)5can be written as [∵According to binomial theorem]

((1 − 2𝑥)5 = 𝐶05 (1)2 − 𝐶1

5 (1)4(2𝑥) + 𝐶25 (1)3(2𝑥)2 − 𝐶3

5 (1)2(2𝑥)3 + 𝐶45 (1) (2𝑥)4 −

𝐶55 (2𝑥)5

= 1 − 5(2𝑥) + 10(4𝑥2) − 10(8𝑥3) + 5(16𝑥4) − (32𝑥5)

= 1 − 10𝑥 + 40𝑥2 − 80𝑥3 + 80𝑥4 − 32𝑥5

2. Expand the expression (2

𝑥−

𝑥

2)

5


𝑛


Solution:

Solution step 1:(2

𝑥−

𝑥

2)

5

Expression (2

𝑥−

𝑥

2)

5 can be written as [∵According to binomial theorem]

(2

𝑥−

𝑥

2)

5

= 𝐶05 (

2

𝑥)

5

(𝑥

2)

0

− 𝐶15 (

2

𝑥)

4

(𝑥

2) + 𝐶2

5 (2

𝑥)

3

(𝑥

2)

2

− 𝐶35 (

2

𝑥)

2

(𝑥

2)

3

+ 𝐶45 (

2

𝑥)

1

(𝑥

2)

4

− 𝐶55 (

2

𝑥)

0

(𝑥

2)

5

= (2

𝑥)

5

− 5 (16

𝑥4) (

𝑥

2) = 10 (

8

𝑥3) (

𝑥2

4) − 10 (

4

𝑥2) (

𝑥3

8) + 5 (

2

𝑥) (

𝑥4

16) −

𝑥5

32


=32

𝑥5−

40

𝑥3+

20

𝑥− 5𝑥 +

5

8𝑥3 −

𝑥3

32

3. Expand the expression (2𝑥 − 3)6


𝑛


Solution:

Solution step 1:(2𝑥 − 3)6

Expression(2𝑥 − 3)6 can be written as [∵According to binomial theorem]

(2𝑥 − 3)6 = 𝐶06 (2𝑥)6(3)0 − 𝐶1

6 (2𝑥)5(3)1 + 𝐶26 (2𝑥)4(3)2 − 𝐶3

6 (2𝑥)3(3)3

+ 𝐶46 (2𝑥)2(3)4 − 𝐶5

6 (2𝑥)(3)5 + 𝐶66 (2𝑥)0(3)6

= 64𝑥6 − 6(32𝑥5)(3) + 15(16𝑥4)(9) − 20(8𝑥3)(27)

+15(4𝑥2)(81) − 6(2𝑥)(243) + 729

= 64𝑥6 − 576𝑥5 + 2160𝑥4 − 4320𝑥3 + 4860𝑥2 − 2916𝑥 + 729

4. Expand the expression (𝑥

3+

1

𝑥)

5.


𝑛


Solution:

Solution step 1:(𝑥

3+

1

𝑥)

5

Expression(𝑥

3+

1

𝑥)

5 can be written as [∵According to binomial theorem]

(𝑥

3+

1

𝑥)

5

= 𝐶50 (

𝑥

3)

5

(1

𝑥)

𝑜

+ 𝐶51 (

𝑥

3)

4

(1

𝑥) + 𝐶5

2 (𝑥

3)

3

(1

𝑥)

2

+ 𝐶53 (

𝑥

3)

2

(1

𝑥)

3

+ 𝐶54 (

𝑥

3)

1

(1

𝑥)

4

+ 𝐶55 (

𝑥

3)

𝑜

(1

𝑥)

5

=𝑥5

243+ (

𝑥4

81)

5

(1

𝑥) + (

𝑥3

27)

10

(1

𝑥2) + (

𝑥2

9)

10

(1

𝑥3) + (

𝑥

3)

5

(1

𝑥4) +

1

𝑥5


=𝑥5

243+

5𝑥3

81+

10𝑥

27+

10

9𝑥+

5

3𝑥3+

1

𝑥5

5. Expand the expression (𝑥 +1

𝑥)

6.


𝑛


Solution:

Solution step 1:(𝑥 +1

𝑥)

6

Expression(𝑥 +1

𝑥)

6can be written as [∵According to binomial theorem]

(𝑥 +1

𝑥)

6

= 𝐶60(𝑥)6 (

1

𝑥)

0

+ 𝐶61(𝑥)5 (

1

𝑥)

1

+ 𝐶62(𝑥)4 (

1

𝑥)

2

+ 𝐶63(𝑥)3 (

1

𝑥)

3

+ 𝐶64(𝑥)2 (

1

𝑥)

4

+ 𝐶65(𝑥) (

1

𝑥)

5

+ 𝐶66(𝑥)0 (

1

𝑥)

6

= 𝑥6 + 6(𝑥3) (1

𝑥) + 15(𝑥4) (

1

𝑥2) + 20(𝑥3) (

1

𝑥3) + 15(𝑥2) (

1

𝑥4) + 6(𝑥) (

1

𝑥3) +

1

𝑥6

= 𝑥6 + 6𝑥4 + 15𝑥2 + 20 +15

𝑥2+

6

𝑥4+

1

𝑥6

6. Using Binomial theorem, evaluate the following (96)3.


𝑛


Solution:

Solution step 1:(96)3

We can write 96 as (100 − 4).

So 963 = (100 − 4)3

Expression (100 − 4)3can be written as [∵According to binomial theorem]

(100 − 4)3 = 𝐶30(100)3(4)0 − 𝐶3

1(100)2(4)1 + 𝐶32(100)(4)2 − 𝐶3

3(100)0(4)3

= (100)3 − 3(100)2(4) + 3(100)(16) − 64


= 1000000 − 120000 + 4800 − 64

= 884736



𝑛


Solution:


We can write 102 as (100 + 2).

So (100 + 2)5 = (100 + 2)5

Expression (100 + 2)5can be written as [∵According to binomial theorem]

(100 + 2)5 = 𝐶50(100)5(2)0 + 𝐶5

1(100)4(2)1 + 𝐶52(100)3(2)3

+ 𝐶53(100)2(2)3 + 𝐶5

4(100)1(2)4 + 𝐶55(100)0(2)5

= (100)5 + 5(100)4 + (2) + 10(100)3(4) + 10(100)2(8) + 5(100)1(16) + 32

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

= 11040808032



𝑛


Solution:


We can write 101 as (100 + 1).

So (101)4 = (100 + 1)4

Expression (100 + 1)4can be written as [∵According to binomial theorem]

(101)4 = 𝐶40(100)4(1)0 + 𝐶4

1(100)3(1) + 𝐶42(100)2(1)2 + 𝐶4

3(100)1(1)3

+ 𝐶44(100)0(1)4

= (100)4 + 4(100)3 + 6(100)2 + 4(100) + 1


= 100000000 + 4000000 + 60000 + 400 + 1

= 104060401



𝑛


Solution:


We can write 99 as (100 − 1).

∴ (99)5 = (100 − 1)5

Expression (100 − 1)5can be written as [∵According to binomial theorem]

= 𝐶50(100)5 − 𝐶5

1(100)4(1) + 𝐶52(100)3(1)2 − 𝐶5

3(100)2(1)3

+ 𝐶54(100) (1)4 + 𝐶5

5(100)0

(1)5

= (100)5 − 5(100)4 + 10(100)3 + 10(100)2 + 5(100) + 1

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

= 10010000500 − 500100001

= 9509900499

10. Using Binomial theorem, indicate which is larger (1.1)10000 or 1000.


𝑛


Solution:

Solution step 1:We can write 1.1 as (1 + 0.1).

∵ (1.1)10000 = (1 + 0.1)10000

Expression (1 + 0.1)10000can be written as [∵According to binomial theorem]

(1 + 0.1)10000 = 𝐶100000(1)10000(0.1)0 + 𝐶10000

1(1)9999(0.1)1 + other positive terms

= 1 + 10000(0.1) + other positive terms


= 1 + 1000 + other positive terms

= 1001 +other positive terms

> 1000

Hence, (1.1)10000 > 1000

11. Find (𝑎 + 𝑏)4 − (𝑎 − 𝑏)4. Hence, evaluate (√3 + √2)4

− (√3 − √2)4

.


𝑛


Solution:

Solution step 1:Expression (𝑎 + 𝑏)4 𝑎𝑛𝑑 (𝑎 − 𝑏)4can be written as [∵According to binomial theorem]

(𝑎 + 𝑏)4 = 𝐶40𝑎4 + 𝐶4

1𝑎3𝑏 + 𝐶42𝑎3𝑏2 + 𝐶4

3𝑎𝑏3 + 𝐶44𝑏4

(𝑎 − 𝑏)4 = 𝐶40𝑎4 + 𝐶4

1𝑎3𝑏 + 𝐶42𝑎3𝑏2 − 𝐶4

3𝑎𝑏3 + 𝐶44𝑏4

∴ (𝑎 + 𝑏)4 + (𝑎 − 𝑏)4 = ( 𝐶40𝑎4 + 𝐶4

1𝑎3𝑏 + 𝐶42𝑎2𝑏2 + 𝐶4

3𝑎𝑏3 + 𝐶44𝑏4)

−( 𝐶40𝑎4 − 𝐶4

1𝑎3𝑏 + 𝐶42𝑎2𝑏2 − 𝐶4

3𝑎𝑏3 + 𝐶44𝑏4)

= 2[ 𝐶41𝑎3𝑏 + 𝐶4

3𝑎𝑏3]

= 2[4𝑎3𝑏 + 4𝑎𝑏3]

= 8𝑎𝑏(𝑎2 + 𝑏2) … . (i)

Step 2: In equation (i) we putting 𝑎 = √3 , 𝑏 = √2.

(√3 + √2 )4

− (√3 − √2)4

= 8(√3)(√2) ((√3)2

+ (√2)2

)

= 8√6(3 + 2)

= 40√6

12. Find (𝑥 + 1)6 + (𝑥 − 1)6. Hence or otherwise evaluate (√2 + 1)6

+ (√2 − 1)6

.


𝑛



Solution:

Solution step 1: Expression (𝑥 + 1)6 and (𝑥 − 1)6can be written as [∵According to binomial theorem]

(𝑥 + 1)6 = 𝐶60𝑥6 + 𝐶6

1𝑥5 + 𝐶62𝑥4 + 𝐶6

3𝑥3 + 𝐶64𝑥2 + 𝐶6

5𝑥 + 𝐶66𝑥0

(𝑥 − 1)6 = 𝐶60𝑥6 − 𝐶6

1𝑥5 + 𝐶62𝑥4 − 𝐶6

3𝑥3 + 𝐶64𝑥2 − 𝐶6

5𝑥 + 𝐶66𝑥0

∴ (𝑥 + 1)6 + (𝑥 − 1)6 = 2[ 𝐶60𝑥6 + 𝐶6

2𝑥4 + 𝐶64𝑥2 + 𝐶6

6]

= 2[𝑥6 + 15𝑥4 + 15𝑥2 + 1] … . (i)

Step 2: In equation (i) we put 𝑥 = √2,

(√2 + 1)6

+ (√2 − 1)6

= 2 [(√2)6

+ 15(√2)4

+ 15(√2)2

+ 1]

= 2[8 + 60 + 30 + 1]

= 2[99]

= 198

13. Show that 9𝑛+1 − 8𝑛 − 9 is divisible by 64, whenever 𝑛 is a positive integer.


𝑛


Solution:

Solution step 1: We can write 9𝑛+1 as (8 + 1)𝑛+1.

Expression (8 + 1)𝑛+1can be written as [∵According to binomial theorem]

So, (8 + 1)𝑛+1 = 𝐶𝑛+10(8)𝑛+1 + 𝐶𝑛+1

1(8)𝑛 + 𝐶𝑛+12(8)𝑛−1+. . . . . + 𝐶𝑛+1

𝑛(8)1 + 𝐶𝑛+1𝑛+1(8)0

⇒ 9𝑛+1 = [ 𝐶𝑛+10(8)𝑛+1 + 𝐶𝑛+1

1(8)𝑛 + 𝐶𝑛+12(8)𝑛−1+. . . . . + 𝐶𝑛+1

𝑛−1(8)2] + (𝑛 + 1)8 + 1

⇒ 9𝑛+1 = 64[ 𝐶𝑛+10(8)𝑛−1 + 𝐶𝑛+1

1(8)𝑛−2 + 𝐶𝑛+12(8)𝑛−3+. . . . . + 𝐶𝑛+1

𝑛−1] + 8𝑛 + 9

Solution Step2: Add both side (−8𝑛 − 9)

So 9𝑛+1 − 8𝑛 − 9 = 64[ 𝐶𝑛+10(8)𝑛−1 + 𝐶𝑛+1

1(8)𝑛−2 + 𝐶𝑛+12(8)𝑛−3+. . . . . + 𝐶𝑛+1

𝑛−1]

⇒ 9𝑛+1 − 8𝑛 − 9 = 64𝑘

where 𝑘 = 𝐶𝑛+10(8)𝑛−1 + 𝐶𝑛+1

1(8)𝑛−2 + 𝐶𝑛+12(8)𝑛−3+. . . . . + 𝐶𝑛+1

𝑛−1 is a natural number.

So, 9𝑛+1 − 8𝑛 − 9 is divisible by 64, whenever 𝑛 is a positive integer.


14. Prove that ∑ 3𝑟 𝐶𝑛𝑟

𝑛

𝑟=0= 4𝑛.

Solution:

Solution step 1: According to binomial theorem,

∑ 𝐶𝑛𝑟

𝑛

𝑟=0

𝑎𝑛−𝑟𝑏𝑟 = (𝑎 + 𝑏)𝑛 … … (𝑖)

By putting 𝑏 = 3 and 𝑎 = 1 in the equation (i)

∑ 𝐶𝑛𝑟

𝑛

𝑟=0

(1)𝑛−𝑟(3)𝑟 = (1 + 3)𝑛

⇒ ∑ 3𝑟. 𝐶𝑛𝑟

𝑛

𝑟=0

= 4𝑛

Hence proved

Exercise 8.2

1. Find the coefficient of𝑥5 in (𝑥 + 3)8.

Hint: 𝑇𝑟+1 = 𝐶𝑛𝑟(𝑎)𝑛−𝑟(𝑏)𝑟

Solution:

Solutions step 1: To get 𝑥5 term coefficient we have to only get the expression of term 𝑥5.

So 𝑇𝑟+1 = 𝐶8𝑟(𝑥)8−𝑟(3)𝑟

Comparing the 𝑥5 and in 𝑇𝑟+1,

8 − 𝑟 = 5

⇒ 𝑟 = 8 − 5

⇒ 𝑟 = 3

So, the coefficient of 𝑥5 is 𝐶83(3)3

=8!

3! 5!× 33


=8 × 7 × 6 × 5!

3! 5!× 27

=8 × 7 × 6

6× 27

= 1512

2. Find the coefficient of 𝑎5𝑏7 in (𝑎 − 2𝑏)12.


Solution: Solutions step 1: To get 𝑎5𝑏7 term coefficient we have to only get the expression of term 𝑎5𝑏7.

So 𝑇𝑟+1 = 𝐶12𝑟(𝑎)12−𝑟(−2𝑏)𝑟

= 𝐶12𝑟(−2)𝑟(𝑎)12−𝑟(𝑏)𝑟

Comparing the 𝑎5𝑏7 and in 𝑇𝑟+1,

⇒ 5 = 12 − 𝑟

⇒ 𝑟 = 7

So, coefficient of 𝑎5𝑏7 is 𝐶127(−2)7

=12!

5! 7!× (−2)7

= −12 × 11 × 10 × 9 × 8 × 7!

5! 7!× 27

=−12 × 11 × 10 × 9 × 8

120× 128

= −101376

3. Write the general term in the expansion of (𝑥2 − 𝑦)6.

Hint: The general term 𝑇𝑟+1 = 𝐶𝑛𝑟𝑎𝑛−𝑟𝑏𝑟

Solution:

Solutions step 1:(𝑟 + 1)term of any expression is called the general term.


𝑇𝑟+1 = 𝐶6𝑟(𝑥2)6−𝑟(−𝑦)𝑟

= (−1)𝑟 𝐶 6𝑟 𝑥12 −2𝑟𝑦𝑟

4. Write the general term in the expansion of (𝑥2 − 𝑦𝑥)12.

Hint: The general term 𝑇𝑟+1 = 𝐶𝑛𝑟𝑎𝑛−𝑟𝑏𝑟

Solution:

Solutions step 1:(𝑟 + 1)term of any expression is called the general term.

𝑇𝑟+1 = 𝐶12𝑟(𝑥2)12−𝑟(−𝑦𝑥)𝑟

= 𝐶12𝑟𝑥24−25(−1)𝑟𝑦𝑟𝑥𝑟

= (−1)𝑟 𝐶12𝑟𝑥24−𝑟𝑦𝑟

5. Find the 4th term in the expansion of (𝑥 − 2𝑦)12.


Solution:

Marks for step 1: According to the binomial (𝑟 + 1) term.

𝑇4 = 𝑇3+1 = 𝐶123(𝑥)12−3(−2𝑦)3

= 𝐶123. (−2)3. 𝑥9. 𝑦3

= −12!

9! 3!. (2)3𝑥9𝑦3

= −−12 × 11 × 10 × 9!

9! 3!. (2)3𝑥9𝑦3

= −1760𝑥9𝑦3

6. Find the 13th term in the expansion of (9𝑥 −1

3√𝑥)

18, 𝑥 ≠ 0



Solution:

Solutions step 1: According to the binomial (𝑟 + 1) term.

𝑇13 = 𝑇12+1 = 𝐶1812. (9𝑥)18−12 (

−1

3√𝑥)

12

= (−1)12. 𝐶1812. (9𝑥)6 (

1

3√𝑥)

12

=18!

6! 2!× 96 × (

1

3)

12

× 𝑥6 ×1

𝑥6

=18!

6! 2![96 = (32)6 = 312]

=18 × 17 × 16 × 15 × 14 × 13 × 2!

6 × 5 × 4 × 3 × 2 × 1 × 12!

= 18564

7. Find the middle terms in the expansions of (3 −𝑥3

6)

7

.


Solution:

Solutions step 1: Total no. of expression in (3 −𝑥3

6)

7

= 8.

So, the middle terms in the expansion will be4𝑡ℎ and 5𝑡ℎ term.

𝑇4 = 𝑇3+1 = 𝐶73(3)7−3 (−

𝑥3

6)

3

= (−1)3 𝐶73(3)4 (

1

6)

3

(𝑥3)3

= − 𝐶73(3)4 (

1

2333)

3

(𝑥9)

= −7!

4! 3!(

3

23) (𝑥9)


= −7 × 6 × 5 × 4!

3! 4!(

3

8) (𝑥9)

= −7 × 6 × 5

3 × 2 × 1× (

3

8) (𝑥9)

= −105

8𝑥9

𝑇5 = 𝑇4+1 = 𝐶74(3)7−4 (−

𝑥3

6)

4

= (−1)4. 𝐶74(3)3 (

1

6)

4

(𝑥3)4

=7!

3! 4!(3)3 (

1

24 × 34) (𝑥12)

=7 × 6 × 5 × 4!

3 × 2 × 1 × 4!× (

1

16 × 3) (𝑥12)

=35

48𝑥12

8. Find the middle terms in the expansions of (𝑥

3+ 9𝑦)

10.


Solution:

Solutions step 1: Total no. of expression in (𝑥

3+ 9𝑦)

10= 11.

So, the middle terms in the expansion will be 6𝑡ℎ term.

𝑇6 = 𝑇5+1 = 𝐶115 (

𝑥

3)

11−5

(9𝑦)5

=11!

5! 6!× (

𝑥6

36)(95𝑦5)

=6! × 7 × 8 × 9 × 10 × 11

1 × 2 × 3 × 4 × 5 × 6!(𝑥6

36)(310𝑦5)

= 462 𝑥634𝑦5


= 37422 𝑥6𝑦5

9. In the expansion of (1 + 𝑥)𝑚+𝑛, prove that the coefficients of 𝑎𝑚 and 𝑎𝑛 are equal.


Solution:

Solutions step 1: Let assume 𝑎𝑚 occurs in the (𝑟 + 1)𝑡ℎ term of the expansion (1 + 𝑎)𝑚+𝑛

So 𝑇𝑟+1 = 𝐶𝑚+𝑛𝑟(1)𝑚+𝑛−𝑟(𝑎)𝑟

Comparing the 𝑎𝑚 and in 𝑇𝑟+1, we get

𝑟 = 𝑚

So, the coefficient of 𝑎𝑚 is

𝐶𝑚+𝑛𝑚 =

(𝑚 + 𝑛)!

𝑚! (𝑚 + 𝑛 − 𝑚)!=

(𝑚 + 𝑛)!

𝑚! 𝑛!… … . (𝑖)

Assume that 𝑎𝑛 occurs in the (𝑘 + 1)𝑡ℎ term of the expansion (1 + 𝑎)𝑚+𝑛

𝑇𝑘+1 = 𝐶𝑚+𝑛𝑘(1)𝑚+𝑛−𝑘(𝑎)𝑘

Comparing the 𝑎𝑛 and in 𝑇𝑘+1

𝑘 = 𝑛

Therefore, the coefficient of 𝑎𝑛 is

𝐶𝑚+𝑛𝑛 =

(𝑚+𝑛)!

𝑛!(𝑚+𝑛−𝑛)!=

(𝑚+𝑛)

𝑛!𝑚! ……………(ii)

From equation (i) and (ii) we can say that the coefficients of 𝑎𝑚 and 𝑎𝑛 in the expansion of (1 + 𝑎)𝑚+𝑛 are equal.

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.


Solution:

Solutions step 1: So, 𝑇𝑟+1 in the expansion of (𝑥 + 1)𝑛 is


𝑇𝑟+1 = 𝐶𝑛𝑟−2(𝑥)𝑛−(𝑟−2)(1)𝑟−2

= 𝐶𝑛𝑟−1𝑥𝑛−𝑟+2 … … . (i)

𝑇𝑟 in the expansion of (𝑥 + 1)𝑛 is

𝑇𝑟 = 𝐶𝑛𝑟−1(𝑥)𝑛−(𝑟−1)(1)𝑟−1

= 𝐶𝑛𝑟−1𝑥𝑛−𝑟+1 … … . (ii)

𝑇𝑟+1 in the expansion of (𝑥 + 1)𝑛 is

𝑇𝑟+1 = 𝐶𝑛𝑟𝑥𝑛−𝑟(1)𝑟

= 𝐶𝑛𝑟𝑥𝑛−𝑟 … … . (iii)

By question, these coefficients are in the ratio 1: 3: 5, we get

𝐶𝑛𝑟−2

𝐶𝑛𝑟−1

=1

3 [by equation (i) and (ii)]

and 𝐶𝑛

𝑟−1

𝐶𝑛𝑟

=3

5[by equation (ii) and (iii)]

=𝐶𝑛

𝑟−2

𝐶𝑛𝑟−1

=𝑛!

(𝑛 − (𝑟 − 2))! (𝑟 − 2!)×

(𝑛 − (𝑟 − 1))! (𝑟 − 1)!

𝑛!

=(𝑟 − 1)! (𝑛 − 𝑟 + 1)!

(𝑟 − 2)! (𝑛 − 𝑟 + 2)!

=(𝑟 − 1)(𝑟 − 2)! (𝑛 − 𝑟 + 1)!

(𝑟 − 2)! (𝑛 − 𝑟 + 2)(𝑛 − 𝑟 + 1)!

=(𝑟 − 1)

𝑛 − 𝑟 + 2

∴(𝑟 − 1)

𝑛 − 𝑟 + 2=

1

3

⇒ 3𝑟 − 3 = 𝑛 − 𝑟 + 2

Or 4𝑟 − 5 = 𝑛 … … … . . (iv)

Also, 𝐶𝑛

𝑟−1

𝐶𝑛𝑟

=𝑛!

(𝑛−(𝑟−1))!(𝑟−1)!×

(𝑛−𝑟)!𝑟!

𝑛!

=(𝑛 − 𝑟)!

(𝑛 − 𝑟 + 1)! (𝑟 − 1)!

=(𝑛 − 𝑟)! 𝑟(𝑟 − 1)!

(𝑛 − 𝑟 + 1)(𝑛 − 𝑟)! (𝑟 − 1)!

=𝑟

(𝑛 − 𝑟 + 1)


∴𝑟

𝑛 − 𝑟 + 1=

3

5

⇒ 5𝑟 = 3𝑛 − 3𝑟 + 3

⇒ 8𝑟 − 3 = 3𝑛 ……………(v)

Putting value of 𝑛 from equation (iv) to equation (v)

So 8𝑟 − 3 = 3(4𝑟 − 5)

⇒ 8𝑟 − 3 = 12𝑟 − 15

⇒ 18 − 3 = 12𝑟 − 8𝑟

⇒ 12 = 4𝑟

or 4𝑟 = 12

⇒ 𝑟 = 3

Putting value of 𝑟 in equation (iv), we get

𝑛 = 4(3) − 5

or 𝑛 = 7

So, 𝑛 = 7 and 𝑟 = 3

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.


Solution:

Solutions step 1: Assume that 𝑥𝑛 occurs in the (𝑟 + 1)𝑡ℎ term of the expansion of (1 + 𝑥)2𝑛

So 𝑇𝑟+1 = 𝐶2𝑛𝑟(1)2𝑛−𝑟(𝑥)𝑟 = 𝐶2𝑛

𝑟𝑥𝑟

Comparing the terms of 𝑥 in 𝑥𝑛 and in 𝑇𝑟+1

So 𝑟 = 𝑛

So, the coefficient of 𝑥𝑛 in the expansion of (1 + 𝑥)2𝑥 is 𝐶2𝑛𝑛

So, 𝐶2𝑛𝑛 =

(2𝑛)!

(2𝑛−𝑛)!𝑛!=

(2𝑛)!

(𝑛!)2 … … … … (i)

Assume that 𝑥𝑛 occurs in the (𝑘 + 1)𝑡ℎ term of the expansion (1 + 𝑥)2𝑛−1

So 𝑇𝑘+1 = 𝐶2𝑛−1𝑘(1)2𝑛−1−𝑘(𝑥)𝑘


= 𝐶2𝑛−1

𝑘 𝑥𝑘

Comparing the terms of 𝑥 in 𝑥𝑛 and in 𝑇𝑘+1

So 𝑘 = 𝑛

So, the coefficient of 𝑥𝑛 in the expansion of (1 + 𝑥)2𝑛−1 is 𝐶2𝑛−1

𝑛

So, 𝐶2𝑛−1

𝑛 =(2𝑛−1)!

(2𝑛−1−𝑛)!𝑛!

=(2𝑛 − 1)!

(𝑛 − 1)! 𝑛!

=2𝑛(2𝑛 − 1)!

2𝑛(𝑛 − 1)! 𝑛!

=2𝑛!

2𝑛! 𝑛!

=(2𝑛!)

2(𝑛!)2… … … (ii)

From equation (i) and (ii)

=1

2( 𝐶2𝑛

𝑛) = 𝐶2𝑛𝑛

⇒ 𝐶2𝑛𝑛 = 2. 𝐶2𝑛−1

𝑛

Thus, the coefficient of 𝑥𝑛 in the expansion of (1 + 𝑥)2𝑥 is twice the coefficient of 𝑥𝑛 in the expansion of (1 + 𝑥)2𝑛−1.

Hence proved.

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


Solution:

Solutions step 1: Assume that 𝑥2 occurs in the (𝑟 + 1)𝑡ℎ term of the expansion (𝑟 + 1)𝑚, we get

𝑇𝑟+1 = 𝐶𝑚𝑟(1)𝑚−𝑟(𝑥)𝑟

= 𝐶𝑚𝑟𝑥𝑟

Comparing the term of 𝑥 in 𝑥2 and in 𝑇𝑟+1 we get 𝑟 = 2

So, the coefficient of 𝑥2 is 𝐶𝑚2


Coefficient of 𝑥2 in the expansion (1 + 𝑥)𝑚 is 6. [∵By question]

∴ 𝐶𝑚2 = 6

⇒𝑚!

(𝑚 − 2)! 2!= 6

⇒𝑚 × (𝑚 − 1) × (𝑚 − 2)!

2(𝑚 − 2)!= 6

⇒ 𝑚(𝑚 − 1) = 6 × 2 = 12

⇒ 𝑚2 − 𝑚 − 12 = 0

⇒ 𝑚2 − 4𝑚 + 3𝑚 − 12 = 0

⇒ 𝑚(𝑚 − 4) + 3(𝑚 − 4) = 0

⇒ (𝑚 + 3)(𝑚 − 4) = 0

⇒ 𝑚 = −3 and 4

Positive value of 𝑚 = 4.

Miscellaneous Exercise

1. Find 𝑎, 𝑏 and 𝑛 in the expansion of (𝑟 + 1)𝑡ℎ if the first three terms of the expansion are 729,7290 and 30375, respectively.

Hint: 𝑇𝑟+1 = 𝐶𝑛𝑟𝑎𝑛−𝑟𝑏𝑟.

Solution:

Solutions step 1: By question 𝑇1 = 729, 𝑇2 = 7290, 𝑇3 = 30375

So, 𝑇1 = 𝐶𝑛0𝑎𝑛𝑏0 = 729

⇒ 𝑎𝑛 = 729 … … … (i)

𝑇2 = 𝐶𝑛1𝑎𝑛−1𝑏1 = 7290

⇒ 𝑛𝑎𝑛−1𝑏 = 7290 … … … (ii)

𝑇3 = 𝐶𝑛2𝑎𝑛−2𝑏2 = 30375

=𝑛(𝑛 − 1)

2𝑎𝑛−2𝑏2 = 30375 … … … . (iii)

By equation (i) and (ii)

𝑎𝑛

𝑛𝑎𝑛−1𝑏=

729

7290 [dividing (i) by (ii)]


⇒𝑎

𝑛𝑏=

1

10

or 𝑎

𝑏=

𝑛

10… … … . (iv)

By equation (ii) and (iii)

𝑛𝑎𝑛+1𝑏

𝑛(𝑛−1)

2𝑎𝑛−2𝑏2

=7290

30375[dividing (ii) by (iii)]

⇒𝑎

(𝑛 − 1)𝑏=

7290

30375 × 2

Put value of 𝑎

𝑏 from equation (iv)

𝑛

10(𝑛 − 1)=

7290

60750=

3

25

⇒ 25𝑛 = 30(𝑛 − 1)

⇒ 25𝑛 = 30𝑛 − 30

⇒ 30 = 30𝑛 − 25𝑛

⇒ 30 = 5𝑛

or 5𝑛 = 30

⇒ 𝑛 = 6

Put value of 𝑛 in equation (i)

So 𝑎6 = 729

⇒ 𝑎6 = 36

⇒ 𝑎 = 3

Put value of 𝑛 in equation (iv)

So 𝑏 =10𝑎

𝑛

⇒ 𝑏 =10 × 3

6

⇒ 𝑏 = 5

So, 𝑎 = 3, 𝑏 = 5 and 𝑛 = 6

2. Find 𝑎 if the coefficients of 𝑥2 and 𝑥3 in the expansion of (3 + 𝑎𝑥)9 are equal.

Hint: 𝑇𝑟+1 = 𝐶𝑛𝑟𝑎𝑛−𝑟𝑏𝑟


Solution:

Solutions step 1: assume that 𝑥2 occurs in the (𝑟 + 1)𝑡ℎ term in the expansion of (3 + 𝑎𝑥)9, we get

𝑇𝑟+1 = 9𝑐𝑟(3)9−𝑟(𝑎𝑥)𝑟

= 9𝑐𝑟(3)9−𝑟𝑎𝑟 · 𝑥𝑟

Comparing the indices of 𝑥 in 𝑥3 and in 𝑇𝑟+1

So 𝑟 = 2

So, coefficient of 𝑥2 is 𝐶92(3)9−2𝑎2 =

9!

2!7!37𝑎2

=9 × 8 × 7!

2 × 1 × 7!37𝑎

= 36 × 37𝑎2

Assume that 𝑥3 occurs in the (𝑘 + 1)𝑡ℎ term in the expansion of (3 + 𝑎𝑥)9

𝑇𝑘+1 = 9𝑐𝑘(3)𝑔−𝑘(𝑎𝑥)𝑘

= 9𝑐𝑘(3)9−𝑘𝑎𝑘𝑥𝑘

Comparing the indices of 𝑥 in 𝑥3 and in 𝑇𝑘+1,

we get 𝑘 = 3

So, the coefficient of 𝑥3 is 𝐶93(3)9−3𝑎3

=9!

3! 6!36𝑎3

=9 × 8 × 7 × 6!

3 × 2 × 1 × 6· 36𝑎3

= 84.36 · 𝑎3

Coefficient of 𝑥2 and 𝑥3 are same [∵by question]

So, 36 · 37𝑎2 = 84 · 36𝑎3

⇒ 84𝑎 = 36 × 3

⇒ 𝑎 =36 × 3

84

𝑎 =3 × 3

7

⇒ 𝑎 =9

7

So, the value of 𝑎 is 9

7.


3. Find the coefficient of 𝑥5 in the product (1 + 2𝑥)6, (1 − 𝑥)7 using binomial theorem.


𝑛


Solution:

Solutions step 1: Expression(1 + 2𝑥)6 and (1 − 𝑥)7,can be written as [∵According to binomial theorem]

(1 + 2𝑥)6 = 𝐶60 + 𝐶6

1(2𝑥) + 𝐶62(2𝑥)2 + 𝐶6

3(2𝑥)3 + 𝐶64(2𝑥)4

+ 𝐶65(2𝑥)5 + 𝐶6

6(2𝑥)6

= 1 + 6(2𝑥) + 15(4𝑥2) + 20(8𝑥3) + 15(16𝑥4) + 6(32𝑥3) + 64𝑥6

= 1 + 12𝑥 + 60𝑥2 + 160𝑥3 + 240𝑥4 + 192𝑥5 + 64𝑥6

(1 − 𝑥)7 = 𝐶70 − 𝐶7

1𝑥 + 𝐶72𝑥2 − 𝐶7

3𝑥3 + 𝐶74𝑥4 − 𝐶7

6𝑥5 + 𝐶76𝑥6 − 𝐶7

7𝑥7

= 1 − 7𝑥 + 21𝑥2 − 35𝑥3 + 35𝑥4 + 21𝑥5 + 7𝑥6 − 𝑥7

∴ (1 + 2𝑥)6(1 − 𝑥)7 = (1 + 12𝑥 + 60𝑥2 + 160𝑥3 + 240𝑥4 + 192𝑥5 + 64𝑥6)(1 − 7𝑥 + 21𝑥2

−35𝑥3 + 35𝑥4 + 21𝑥5 + 7𝑥6 + 𝑥7)

So the terms containing 𝑥5 are

1(−21𝑥5) + (12𝑥)(35𝑥4) + (60𝑥2)(−35𝑥3) + (160𝑥3)(21𝑥2) + (240𝑥4)(−7𝑥)

+(192𝑥5)(1)

= 171𝑥5

So the coefficient of 𝑥5 is 171.

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 ∑ 𝐶𝑛𝑟

𝑛


Solution:

Solutions step 1:a can be written as 𝑎 − 𝑏 + 𝑏

So 𝑎𝑛 = [(𝑎 − 𝑏) + 𝑏]𝑛

= 𝐶𝑛0(𝑎 − 𝑏)𝑛 + 𝐶𝑛

1(𝑎 − 𝑏)𝑛−1𝑏 + 𝐶𝑛2(𝑎 − 𝑏)𝑛−2𝑏2 + ⋯ + 𝐶𝑛

𝑛−1(𝑎 − 𝑏)𝑏𝑛−1

+ 𝐶𝑛𝑛𝑏𝑛


⇒ 𝑎𝑛 = (𝑎 − 𝑏)[ 𝐶𝑛0(𝑎 − 𝑏)𝑛−1 + 𝐶𝑛

1(𝑎 − 𝑏)𝑛−2𝑏+. . . . . + 𝐶𝑛𝑛−1𝑏𝑛−1] + 𝑏𝑛

⇒ 𝑎𝑛 − 𝑏𝑛 = (𝑎 − 𝑏). 𝑘 … … . (i)

where 𝑘 = [ 𝐶𝑛0(𝑎 − 𝑏)𝑛−1 + 𝐶𝑛

1(𝑎 − 𝑏)𝑛−2𝑏+. . . . . + 𝐶𝑛𝑛−1𝑏𝑛−1] is a natural number.

By using equation (i)

= (𝑎 − 𝑏).𝑘

𝑎 − 𝑏

= 𝑘

Hence, it is proved that (𝑎 − 𝑏) is a factor of (𝑎𝑛 − 𝑏𝑛).

5. Evaluate (√3 + √2)6

− (√3 + √2)6

.

∑ 𝐶𝑛𝑟

𝑛

𝑟=0

𝑎𝑛−𝑟𝑏𝑟 = (𝑎 + 𝑏)𝑛

Solution:

Solutions step 1:(𝑎 + 𝑏)6 = 𝐶60𝑎6 − 𝐶6

1𝑎5𝑏 + 𝐶62𝑎4𝑏2 − 𝐶6

3𝑎3𝑏3 + 𝐶64𝑎2𝑏4 − 𝐶6

5𝑎𝑏5 +

𝐶66𝑏6 … … . . (i)

(𝑎 − 𝑏)6 = 𝐶60𝑎6 − 𝐶6

1𝑎5𝑏 + 𝐶62𝑎4𝑏2 − 𝐶6

3𝑎3𝑏3 + 𝐶64𝑎2𝑏4 − 𝐶6

5𝑎𝑏5

+ 𝐶66𝑏6 … … . . (ii)

On subtracting equation (ii) from (i), we get

(𝑎 + 𝑏)6 − (𝑎 − 𝑏)6 = 2[ 𝐶61𝑎5𝑏 − 𝐶6

3𝑎3𝑏3 + 𝐶65𝑎𝑏3]

= 2[6𝑎3𝑏 + 20𝑎3𝑏3 + 6𝑎𝑏3]

Step 2: Putting 𝑎 = √3 ,𝑏 = √2

(√3 + √2)6

− (√3 − √2)6

= 2 [6(√3)5

(√2) + 20(√3)3

(√2)3

+ 6(√3)(√2)5

]

= 2[54√6 + 120√6 + 24√6]

= 2 × 198√6

= 396√6


6. Evaluate (𝑎2 + √𝑎2 − 1)4

+ (𝑎2 − √𝑎2 − 1)4

.

∑ 𝐶𝑛𝑟

𝑛

𝑟=0

𝑎𝑛−𝑟𝑏𝑟 = (𝑎 + 𝑏)𝑛

Solution:

(𝑥 + 𝑦)4 = 𝐶40𝑥4 + 𝐶4

1𝑥3𝑦 + 𝐶42𝑥2𝑦2 + 𝐶4

3𝑥 𝑦3 + 𝐶44𝑦4 … … (i)

(𝑥 − 𝑦)4 = 𝐶40𝑥4 − 𝐶4

1𝑥3𝑦 + 𝐶42𝑥2𝑦2 − 𝐶4

3𝑥𝑦3 + 𝐶44𝑦4 … … (ii)

On adding equation (i) and (ii), we get

(𝑥 + 𝑦)4 + (𝑥 − 𝑦)4 = 2[ 𝐶40𝑥4 + 𝐶4

2𝑥4𝑦2 + 𝐶44𝑦4]

= 2[𝑥4 + 6𝑥2𝑦2 + 𝑦4]

Step 2: Putting 𝑥 = 𝑎2, 𝑦 = √𝑎2 − 1

(𝑎2 + √𝑎2 − 1)4

+ (𝑎2 − √𝑎2 − 1)4

= 2 [(𝑎2)4 + 6(𝑎2)2 (√𝑎2 − 1)2

+ (√𝑎2 − 1)4

]

= 2[𝑎8 + 6𝑎4(𝑎2 − 1) + (𝑎2 − 1)2]

= 2[𝑎8 + 6𝑎4 − 6𝑎4 + 𝑎4 + 2𝑎2 + 1]

= 2[𝑎8 + 6𝑎6 − 5𝑎4 − 2𝑎2 + 1]

= 2𝑎8 + 12𝑎6 − 10𝑎4 − 4𝑎2 + 2

7. Find an approximations of (0.99)5 using the first three terms of its expansion.


𝑛


Solution:

Solutions step 1:We can write 0.99 can as (1 − 0.01).

∴ (0.99)5 = (1 − 0.01)5

So, (1 − 0.01)5 = 𝐶50(1)5 − 𝐶5

1(1)4(0.01) + 𝐶52(1)3(0.01)2

(we take value as close as possible)


= 1 − 5(0.01) + 10(0.01)2

= 1 − 0.05 + 0.001

= 1.001 − 0.05

= 0.951

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

8. Find 𝑛, if the ratio of the term from the beginning to the fifth term from the end in the

expansion of (√24

+1

√34 )

𝑛 is √6: 1.


𝑛


Solution:

Solutions step 1: Fifth term from beginning = 𝐶𝑛4𝑎𝑛−4𝑏4

Fifth term from end = 𝐶𝑛𝑛−4𝑎4𝑏𝑛−4

Putting 𝑎 = √24

,𝑏 =1

√34

Fifth term from the beginning is 𝐶𝑛4(√2

4)

𝑛−4(

1

√34 )

4

Fifth term from the end is 𝐶𝑛𝑛−4(√2

4)

4(

1

√34 )

𝑛−4

𝐶𝑛𝑛( √2

4)

𝑛−4(

1

√34 )4

𝐶𝑛𝑛−4( √2

4)

4(

1

√34 )𝑛−4 =

√6

1 [By question]

⇒(√2

4)

𝑛−4(√3

4)

𝑛−4

(√24

)4

(√34

)4 =

√6

1[ 𝐶𝑛

𝑟 = 𝐶𝑛𝑛−𝑟]

⇒ (√24

)𝑛−8

(√34

)𝑛−8

= √6

⇒ (√64

)𝑛−8

= √6

⇒ (6)𝑛−8

4 = (6)1

2

Comparing powers on both sides

So 𝑛−8

4=

1

2


⇒ 𝑛 − 8 = 2

⇒ 𝑛 = 10

So, the value of 𝑛 is 10.

9. Expand using Binomial theorem (1 +𝑥

2−

2

𝑥)

4, 𝑥 ≠ 0.


𝑛


Solution:

Solutions step 1:We can write (1 +𝑥

2−

2

𝑥)

4 as [(1 +

𝑥

2) −

2

𝑥]

4.

Expression[(1 +𝑥

2) −

2

𝑥]


= 𝐶40 (1 +

𝑥

2)

4

− 𝐶41 = (1 +

𝑥

2)

3

(2

𝑥)

1

+ 𝐶42 = (1 +

𝑥

2)

2

(2

𝑥)

2

− 𝐶43

(1 +𝑥

2) (

2

𝑥)

3

+ 𝐶44 = (

2

𝑥)

4

= (1 +𝑥

2)

4

− 4 (1 +𝑥

2)

3

(2

𝑥) + 6 (1 +

𝑥

2)

2

(4

𝑥2) − 4 (1 +

𝑥

2) (

8

𝑥3) +

16

𝑥4

= (1 +𝑥

2)

4

−8

𝑥(1 +

𝑥

2)

3

+24

𝑥2+

24

𝑥+ 6 −

32

𝑥3−

16

𝑥2+

16

𝑥4

= (1 +𝑥

2)

4

−8

𝑥(1 +

𝑥

2)

3

+8

𝑥2+

24

𝑥+ 6 −

32

𝑥3+

16

𝑥4… … (i)

Again, by using binomial theorem,

= (1 +𝑥

2)

4

− 𝐶40(1)4 + 𝐶4

1(1)3 (𝑥

2) + 𝐶4

2(1)2 (𝑥

2)

2

+ 𝐶43(1)1 (

𝑥

2)

3

+ 𝐶43 (

𝑥

2)

4

= 1 + 4 (𝑥

2) + 6 (

𝑥2

4) + 4 (

𝑥3

8) + (

𝑥4

16)

= 1 + 2𝑥 +3

2𝑥2 +

𝑥3

2+

𝑥4

16… … (ii)

(1 +𝑥

2)

3

= 𝐶30(1)3 + 𝐶3

1(1)2 (𝑥

2) + 𝐶3

2(1) (𝑥

2)

2

+ 𝐶33 (

𝑥

2)

3

= 1 + 3 (𝑥

2) + 3 (

𝑥2

4) + (

𝑥3

8)


= 1 +3𝑥

2+

3𝑥2

4+

𝑥3

8… … (iii)

From equation (i), (ii) and (iii)

[(1 +𝑥

2) − (

2

𝑥)]

4

= 1 + 2𝑥 +3𝑥2

2+

𝑥3

2+

𝑥4

16−

8

𝑥(1 +

3𝑥

2+

3𝑥2

4+

𝑥3

8) +

8

𝑥2+

24

𝑥+ 6 −

32

𝑥3

+16

𝑥4

= 1 + 2𝑥 +3𝑥2

2+

𝑥3

2+

𝑥4

16−

8

𝑥− 12 − 6𝑥 − 𝑥2 +

8

𝑥2+

24

𝑥+ 6 −

32

𝑥3+

16

𝑥4

=16

𝑥+

8

𝑥2−

32

𝑥3+

16

𝑥4− 4𝑥 +

𝑥2

2+

𝑥3

2+

𝑥4

16− 5

10. Find the expansion of (3𝑥2 − 2𝑎𝑥 + 3𝑎2)3 using binomial theorem.


𝑛


Solution:

Solutions step 1: We can write (3𝑥2 − 2𝑎𝑥 + 3𝑎2)3 as ((3𝑥2 − 2𝑎𝑥) + 3𝑎2)3

Expression[(1 +𝑥

2) −

2

𝑥]


(3𝑥2 − 2𝑎𝑥 + 3𝑎2)3

= 𝐶30(3𝑥2 − 2𝑎𝑥)3 + 𝐶3

1(3𝑥2 − 2𝑎𝑥)2(3𝑎2) + 𝐶32(3𝑥2 − 2𝑎𝑥)(3𝑎2)2 + 𝐶3

3(3𝑎2)3

= (3𝑥2 − 2𝑎𝑥)3 + 3(9𝑥4 + 4𝑎2𝑥2 − 12𝑎𝑥3)(3𝑎2) + 3(3𝑥2 − 2𝑎𝑥)(9𝑎4) + (27𝑎6)

= (3𝑥2 − 2𝑎𝑥)3 + 81𝑎2𝑥4 + 36𝑎4𝑥2 − 108𝑎3𝑥3 + 81𝑥2𝑎4 − 54𝑎4𝑥 + 27𝑎6 … … (i)

Again, by using binomial theorem, we get

= (3𝑥2 − 2𝑎𝑥)3 = 𝐶30(3𝑥2)3 − 𝐶3

1(3𝑥2)2(2𝑎𝑥) + 𝐶32(3𝑥2)2(2𝑎𝑥)2 − 𝐶3

3(2𝑎𝑥)3

= 27𝑥6 + 3(9𝑥4)(2𝑎𝑥) + 3(3𝑥2)(4𝑎2𝑥2) − (8𝑎3𝑥3)

= 27𝑥6 − 54𝑎𝑥5 + 36𝑎2𝑥4 − 8𝑎3𝑥3 … … (ii)

By adding equation (i) and (ii)

= 27𝑥6 − 54𝑎𝑥5 + 36𝑎2𝑥4 − 8𝑎3𝑥3 + 81𝑎2𝑥4 + 117𝑎4𝑥2 − 108𝑎3𝑥3 + 81𝑎4𝑥2 − 54𝑎5𝑥

+27𝑎6


= 27𝑥6 − 54𝑎𝑥5 + 117𝑎2𝑥4 − 116𝑎3𝑥3 + 117𝑎4𝑥2 − 54𝑎5𝑥 + 27𝑎6

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Class XI CBSE-Mathematics Binomial Theorem · 2019-11-27 · Class–XI–CBSE-Mathematics Binomial Theorem 5 243 + 5 3 81 + 10 27 + 10 9 + 5 3 3 1 5 5. Expand the expression ( +1