NCERT Exemplar: Binomial Theorem | Mathematics (Maths) Class 11 - Commerce PDF Download

Q.1. Find the term independent of x, x ≠ 0, in the expansion of.
Ans.
General Term T_r+1 = ⁿC_rx^n–ry^r

For getting the term independent of x,
30 – 3r = 0 ⇒ r = 10
On putting the value of r in the above expression, we get

Hence, the required term =

Q.2. If the term free from x in the expansion ofis 405, find the value of k.
Ans. 
The given expression is 
General term T_r+1 = ⁿC_r x^n-r y^r


For getting term free from x,
⇒ r = 2
On putting the value of r in the above expression, we get
¹⁰C₂ (–K)²
According to the condition of the question, we have  
  
⇒ 45K² = 405 ⇒
∴ K = ± 3
Hence, the value of K = ±3

Q.3. Find the coefficient of x in the expansion of (1 – 3x + 7x²) (1 – x)¹⁶.
Ans.
The  given expression is (1 – 3x + 7x²) (1 – x)¹⁶ = (1 – 3x + 7x²) [¹⁶C₀(1)¹⁶(–x)⁰ + ¹⁶C₁(1)¹⁵ (–x) + ¹⁶C₂(1)¹⁴ (–x)² + …]
= (1 – 3x + 7x²) (1 – 16x + 120x² …)
Collecting the term containing x, we get –16x – 3x = – 19x
Hence, the coefficient of x = –19

Q.4. Find the term independent of x in the expansion of,
Ans.
Given expression is
General term T_r+1 = ⁿC_rx^n–ry^r

For getting a term independent of x, put 15 – 3r = 0 ⇒ r = 5
∴ The required term is ¹⁵C₅(3)^15-5(-2)⁵

Hence, the required term = –3003 (3)¹⁰ (2)⁵

Q.5. Find the middle term (terms) in the expansion of
(i)
(ii) 
Ans.
(i) Given expression is
Number of terms = 10 + 1 = 11 (odd)
∴ Middle term =
General Term T_r+1 = ⁿC_rx^n–r y^r

Hence, the required middle term = – 252
(ii) Given expression is
Number of terms = 9 + 1 = 10 (even)
∴ Middle terms are
= 5^th term and (5 + 1) = 6^th term
General  Term T_r+1 = ⁿC_rx^n–ry^r

                       
Now, T₆ = T₅₊₁ =


Hence, the required middle terms are

Q.6.  Find the coefficient of x¹⁵ in the expansion of (x – x²)¹⁰.
Ans.
The given expression is (x – x²)¹⁰
General Term T_r+1 = ⁿC_rx^n–ry^r
= ¹⁰C_r(x)^10–r (–x²)^r = ¹⁰C_r(x)^10–r (–1)^r × (x²)^r
= (-1)^r × ¹⁰C_r(x)^10-r+2r = (- 1)^r × ¹⁰C_r(x)^10+r
To find the coefficient of x¹⁵, Put 10 + r = 15 ⇒ r = 5
∴ Coefficient of x15 = (-1)^{5 10}C₅ = - ¹⁰C₅ = - 252
Hence, the required coefficient = - 252

Q.7. Find the coefficient ofin the expansion of
Ans.
The given expression is
General Term T_r+1 = ⁿC_rx^n–ry^r  = ¹⁵C_r(4)^15-r
= ¹⁵C_r (x)^{60 – 4r} (- 1)^r

To find the coefficient ofPut 7r – 60 = 17
⇒ 7r  = 60 + 17 ⇒ 7r = 77
∴ r = 11
Putting the value of r in the above expression, we get   

Hence, the coefficient of

Q.8. Find the sixth term of the expansionf the binomial coefficient of the third term from the end is 45.
[Hint: Binomial coefficient of third term from the end = Binomial coefficient of third term from beginning = ⁿC₂.]
Ans.
The given expression is ( y^1/2 + x^1/3 )ⁿ, since the binomial
coefficient of third term from the end = 
Binomial coefficient of  third term from the beginning = ⁿC₂
∴ ⁿC₂ = 45
⇒ ⇒ n² – n = 90
⇒ n² – n – 90 = 0 ⇒ n² – 10n + 9n – 90 = 0
⇒ n(n – 10) + 9(n – 10) ⇒ (n – 10) (n + 9) = 0
⇒ n = 10, n = - 9 ⇒ n = 10, n ≠ - 9
So, the given expression becomes (y^1/2 + x^1/3)¹⁰
Sixth term is this expression
T₆ = T₅₊₁ = ¹⁰C₅(y^1/2)^10–5 (x^1/3)⁵ = ¹⁰C₅y^5/2 × x^5/3
= 252 y^5/2x^5/3
Hence, the required term = 252 y^5/2 . x^5/3

Q.9. Find the value of r, if the coefficients of (2r + 4)^th and (r – 2)^th terms in the expansion of (1 + x)¹⁸ are equal.
Ans.
General Term T_r+1 = ⁿC_r x^n–ry^r
For coefficient of (2r + 4)^th term, we have
T_2r+4 = T_2r+3+1 = ¹⁸C_2r+3 (1)^18–2r–3 × x^2r+3
∴ Coefficient of  (2r + 4)^th term = ¹⁸C_2r+3
Similarly, T_r-2 = T_r-3+1 = ¹⁸C_r-3 (1)^18-r+3 × x^r-3
∴ Coefficient of (r - 2)^th term = ¹⁸C_r3
As per the condition of the questions, we have ¹⁸C_2r+3 = ¹⁸C_r-3 
⇒ 2r + 3 + r - 3  = 18 ⇒ 3r = 18 ⇒ r = 6

Q.10. If the coefficient of second, third and fourth terms in the expansion of (1 + x)²ⁿ are in A.P. Show that 2n² – 9n + 7 = 0.
Ans.
Given expression = (1 + x )²ⁿ
Coefficient of second term = ²ⁿC₁
Coefficient of third term = ²ⁿC₂
and coefficient of fourth term = ²ⁿC₃
As the given condition, ²ⁿC₁, ²ⁿC₂ and ²ⁿC₃ are in A.P.
∴ ²ⁿC₂ – ²ⁿC₁ = ²ⁿC₃ – ²ⁿC₂ 
⇒ 2 × ²ⁿC₂ = ²ⁿC₁ + ²ⁿC<sub>3


⇒
⇒</sub>
⇒ 12n – 6 = 6 + 4n² – 4n – 2n + 2
⇒ 4n² –  6n – 12n + 2 + 12 = 0
⇒ 4n² – 18n  + 14 = 0
⇒ 2n² – 9n + 7 = 0
Hence proved.

Q.11. Find the coefficient of x⁴ in the expansion of  (1 + x + x² + x³)¹¹.
Ans.
 Given expression is (1 + x + x² + x³)¹¹
= [(1 + x) + x² (1 + x)]¹¹  = [(1 + x) (1 + x²)]¹¹
= (1 + x)¹¹ × (1 + x²)¹¹
Expanding the above expression, we get (¹¹C₀ + ¹¹C₁x + ¹¹C₂x² + ¹¹C₃x³ + ¹¹C₄x⁴ + …) × (¹¹C₀ + ¹¹C₁x² + ¹¹C₂x⁴ + )
=(1 + 11x + 55x² + 165x³ + 330x⁴ …) × (1 + 11x² + 55x⁴ + &)
Collecting the terms containing x⁴, we get
(55 + 605 + 330)x⁴ = 990x⁴
Hence, the coefficient of x⁴ = 990

Long AnswerType
Q.12. If p is a real number and if the middle term in the expansion of is 1120, find p.
Ans.
Given expression is 
Number of terms = 8 + 1 = 9 (odd)
∴ Middle term =  term = 5th term  
∴

Now  ⁸C₄P⁴ = 1120 ⇒
⇒ 70P⁴ = 1120
⇒
Hence, the required value of P = ±2    

Q.13. Show that the middle term in the expansion ofis

Ans.
Given expression is
Number of terms = 2n + 1 (odd)
∴ Middle term =  th i.e.,    (n + 1)^th term

  Hence, the middle term =

Q.14. Find n in the binomial if the ratio of 7th term from the beginning to the 7th term from the end is 
Ans.
 The given expression is 
 
General Term T_r+1 = ⁿC_rx^n–ry^r

7th term from the end = (n – 7 + 2)^th term from the beginning = (n – 5)^th term from the beginning
So,

According to the question, we get



Hence, the required value of n is 9.

Q.15. In the expansion of (x + a)ⁿ if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that
(i) O² – E² = (x² – a²)ⁿ
(ii) 4OE = (x + a)²ⁿ – (x – a)²ⁿ
Ans.
Given expression is (x + a)ⁿ
(x + a)ⁿ = ⁿC₀xⁿa⁰ + ⁿC₁x^n–1a + ⁿC₂x^n–2a² + ⁿC₃x^n–3a³ + … + ⁿC_naⁿ
Sum of odd terms,
O = ⁿC₀ xⁿ + ⁿC₂ x^{n- 2} a²+ ⁿC₄ x^{n- 4} a⁴+ ...
and the sum of even terms 
E = ⁿC₁ x^n-1 × a + ⁿC₃ x^{n- 3}a³+ ⁿC₅ x^{n- 5}a⁵+ ...
Now (x + a)ⁿ = O + E ...(i)
Similarly (x - a)ⁿ = O - E ...(ii)
Multiplying eq. (i) and eq. (ii), we get,
(x + a)ⁿ (x – a)ⁿ = (O + E) (O – E)
⇒ (x² – a²)ⁿ = O² – E²
Hence O² – E² = (x² – a²)ⁿ
(ii) 4OE = (O + E)² – (O – E)²
= [(x + a)ⁿ]² – [(x – a)ⁿ]²
= [x + a]²ⁿ – [x – a]²ⁿ
Hence, 4OE = (x + a)²ⁿ – (x – a)²ⁿ

Q.16. If x^p occurs in the expansion of prove that its coefficient is 
Ans.
Given expression is
General terms, T_r+1 = ⁿC_rx^n–ry^r

If x^p occurs in
then 4n – 3r = p ⇒ 3r = 4n - p
⇒
∴ Coefficient of x^p =

Hence, the coefficient of x^p =

Q.17. Find the term independent of x in the expansion of (1 + x + 2x³) 
Ans.
Given expression is 
Let us consider
General Term T_r+1 = ⁿC_rx^n–ry^r


So, the general term in the expansion of

For getting the term independent of x,
Put 18 – 3r = 0, 19 – 3r = 0 and 21 – 3r = 0, we get

The possible value of r are 6 and 7
∴ The term independent of x is

Hence, the required term =

Objective Type Questions
Q.18. The total number of terms in the expansion of (x + a)¹⁰⁰ + (x – a)¹⁰⁰ after simplification is
(a) 50
(b) 202
(c) 51
(d) none of these
Ans. (c)
Solution.
Number of terms in the expansion of (x + a)¹⁰⁰ = 101
Number of terms in the expansion of (x – a)¹⁰⁰ = 101
Now 50 terms of expansion will cancel out with negative 50 terms of (x – a)¹⁰⁰ So, the remaining 51 terms of first expansion will be added to 51 terms of other.
Therefore, the number of terms = 51
Hence, the correct option is (c).  

Q.19. Given the integers r > 1, n > 2, and coefficients of (3r)^th and  (r + 2)^nd terms in the binomial expansion of (1 + x)²ⁿ are equal, then
(a) n = 2r
(b) n = 3r
(c) n = 2r + 1
(d) none of these
Ans. (a)
Solution.
Given that r > 1 and n > 2
then T_3r = T_3r–1+1 = ²ⁿC_3r–1 × x^3r–1
and T_r+2 = T_{r +1+1} = ²ⁿC_{r +1} x^{r +1}
As per the question, we have   
As  per the question, we have
²ⁿC_3r–1 = ²ⁿC_r+1
⇒ 3r – 1 + r + 1 = 2n   [∴ ⁿC_p = ⁿC_q ⇒ n = p + q]
⇒ 4r = 2n
n = 2r
Hence, the correct option is (a).

Q.20. The two successive terms in the expansion of (1 + x)²⁴ whose coefficients are in the ratio 1:4 are
(a) 3rd and 4th
(b) 4th and 5th
(c) 5th and 6th
(d) 6th and 7th
[Hint: ⇒ 4r + 4 = 24 – 4 ⇒ r = 4 ]
Ans. (c)
Solution.
 Let r^th and (r + 1)^th be two successive terms in the  expansion  (1 + x)²⁴
∴ T_r+1 = ²⁴C_r × x^r
T_r+2 = T_r+1+1 = ²⁴C_r+1x^r+1
As per the question, we have


⇒ 5r = 20 ⇒ r = 4
∴ T₄₊₁ = T₅ and T₄₊₂ = T₆
Hence, the correct option is (c).

Q.21. The coefficient of xⁿ in the expansion of (1 + x)²ⁿ and (1 + x)^{2n – 1} are in the ratio.
(a) 1 : 2
(b) 1 : 3
(c) 3 : 1
(d) 2 : 1
[Hint : ²ⁿC_n : ^{2n – 1}C_n
Ans. (d)
Solution.
General Term T_r+1 = ⁿC_r x^n–r y^r
In the expansion of (1 + x)²ⁿ, we get T_r+1 = ²ⁿC_r x^r
To get the coefficient of xⁿ, put r = n
∴ Coefficient of xⁿ = ²ⁿC_n
In the expansion of (1 + x)^2n-1, we get T_r+1 = ^2n-1C_rx^r
∴ Coefficient of xⁿ is 2^n-1C_n-1
The required ratio is


Hence, the correct option is (d).

Q.22. If the coefficients of 2^nd, 3^rd and the 4^th terms in the expansion of (1 + x)n are in A.P., then value of n is
(a) 2
(b) 7
(c) 11
(d) 14
[Hint: ²ⁿC₂ = ⁿC₁ + ⁿC₃ ⇒ n² – 9n + 14 = 0 ⇒ n = 2 or 7]
Ans. (b)
Solution.
Given expression is (1 + x)ⁿ 
(1 + x)ⁿ = ⁿC₀ + ⁿC₁x + ⁿC₂x² + ⁿC₃x³ + … ⁿC_nxⁿ Here, coefficient of 2nd term = ⁿC₁
Coefficient of 3rd term = ⁿC₂
and coefficient of 4th term = ⁿC₃
Given that ⁿC₁, ⁿC₂ and ⁿC₃ are in A.P.     
∴ 2 . ⁿC₂ = ⁿC₁ + ⁿC₃

⇒ 6n – 6 = 6 + n² – 3n + 2
⇒ n² – 3n – 6n + 14 = 0 ⇒ n² – 9n + 14 = 0
⇒ n² – 7n – 2n + 14 = 0 ⇒ n(n – 7) – 2(n – 7) = 0
⇒ (n – 2) (n – 7) = 0 ⇒ n = 2, 7 ⇒ n = 7
whereas n  = 2 is not possible
Hence, the correct option is (b). 

Q.23. If A and B are coefficient of xⁿ in the expansions of (1 + x)²ⁿ and (1 + x)^{2n – 1} respectively, thenequals
(a) 1
(b) 2 
(c) 1/2
(d) 1/n

Ans. (b)
Solution.
Given expression is (1 + x)²ⁿ
T_r+1 = ²ⁿC_rx^r
∴ Coefficient of xⁿ = ²ⁿC_n = A (Given)
In the expression (1 + x)^2n–1
T_r+1 = ^2n–1C_rx^r
∴ Coefficient of xⁿ = ^2n-1C_n = B (Given)
So [from Q. no. 21]
Hence, the correct option is (b).

Q.24. If the middle term ofis equal tothen value of x is
(a)
(b)
(c) 
(d)

Ans. (c)
Solution.
Given expression is
Number of terms = 10 + 1 = 11 odd
∴ Middle term =term = 6th term

⇒
⇒
∴
Hence, the correct option is (c).

Fill in the blanks
Q.25. The largest coefficient in the expansion of (1 + x)³⁰ is  _____.
Ans.
Here n = 30 which is even
∴ the largest coefficient in (1 + x)ⁿ = ⁿC_n/2 
So, the largest coefficient in (1 + x)³⁰ = ³⁰C₁₅ 
Hence, the value of the filler is ³⁰C₁₅.

Q.26. The number of terms in the expansion of (x + y + z)ⁿ ________.
[Hint: (x + y + z)ⁿ = [x + (y + z)]ⁿ]
Ans.
The expression (x + y + z)ⁿ can be written a [x + (y + z)]ⁿ
∴ [x + (y + z)]ⁿ = ⁿC₀xⁿ (y + z)⁰ + ⁿC₁(x)^n–1 (y + z)
+ ⁿC₂(x)^n–2 (y + z)² + … + ⁿC_n(y + z)ⁿ
∴ Number of terms 1 + 2 + 3 + 4 + & (n + 1)

Hence, the value of the filler is   

Q.27. In the expansion of the value of constant term is_______.
Ans.
Let T_r+1 be the constant term in the expansion of

For getting constant term, 32 – 4r = 0
⇒ = r = 8
∴ T_r + 1 = (– 1)⁸ . ¹⁶C₈ = ¹⁶C₈
Hence, the value of the filler is ¹⁶C₈.

Q.28. If the seventh terms from the beginning and the end in the expansion of
are equal, then n equals _____.

Ans.
The given expansion is
Now the T₇ from the end = T₇ from the beginning in

As per the questions, we get


Hence, the value of the filler is 12.  

Q.29. The coefficient of a^{– 6} b⁴ in the expansion of is _________.

Ans.
The given expansion is from a^–6b⁴, we can take r = 4


Hence, the value of the filler =.

Q.30. Middle term in the expansion of (a³ + ba)²⁸ is _________ .
Ans.
Number of term in the expansion (a³ + ba)²⁸ = 28 + 1 = 29 (odd)
∴ Middle term =  
∴ T₁₅ = T₁₄₊₁ = ²⁸C₁₄ (a³)^28–14 × (ba)¹⁴ = ²⁸C₁₄ (a)⁴² × b¹⁴ × a¹⁴
= ²⁸C₁₄a⁵⁶b¹⁴
Hence, the value of the filler is ²⁸C₁₄ a⁵⁶ b¹⁴.  

Q.31. The ratio of the coefficients of xp and xq in the expansion of (1 + x)^{p + q} is_________ [Hint: ^{p + q}C_p = ^{p + q}C_q]
Ans.
 Given expansion is (1 + x)^p+q
T_r+1 = ^{p + q}C_r x^r
Put r = p = ^{p + q}C_px^p
∴ the  coefficient of x^p = ^{p + q}C_p
Similarly, coefficient of x^q = ^{p + q}C_q

and
So, the ratio is 1 : 1.

Q.32. The position of the term independent of x in the expansion of 
  is _______.
Ans.
The given expansion is  
  

For independent of x, we get 
   
10 – 5r = 0
r = 2
So, the position of the term independent of x is 3rd term.
Hence, the value of the filler is Third term

Q.33. If 2515 is divided by 13, the reminder is _________ .
Ans.
Let 25¹⁵ = (26– 1)¹⁵
= ¹⁵C₀ (26)¹⁵  (- 1)⁰ + ¹⁵C₁ (26)¹⁴(- 1)¹ + ¹⁵C₂ (26)¹³  (- 1)² + … + ¹⁵C₁₅(- 1)¹⁵
= 26¹⁵ - 15(26)¹⁴ + L - 1 - 13 + 13
= 26¹⁵ – 15 × (26)¹⁴ + … – 13 + 12
= 13λ + 12
∴ The remainder = 12
Hence, the value of the filler is 12.

True or False.
Q.34. The sum of the series
Ans.
= ²⁰C₀ + ²⁰C₁ + ²⁰C₂ + ²⁰C₃ + … + ²⁰C₁₀

= ²⁰C₀ + ²⁰C₁ + … + ²⁰C₁₀ + ²⁰C₁₁ + … + ²⁰C₂₀ – (²⁰C₁₁+ … + ²⁰C₂₀)
= 220 – (²⁰C₁₁ + … + ²⁰C₂₀)
Hence, the given statement is False.

Q.35. The expression 7⁹ + 9⁷ is divisible by 64.
Hint: 7⁹ + 9⁷ = (1 + 8)⁷ – (1 – 8)⁹
Ans.
7⁹ + 9⁷ = (1 + 8)⁷ – (1 – 8)⁹ = [⁷C₀ + ⁷C₁ × 8 + ⁷C₂(8)² + ⁷C₃(8)³ + & + ⁷C₇(8)⁷] – [⁹C₀ – ⁹C₁8 + ⁹C₂(8)² – ⁹C₃(8)³ + … ⁹C₉(8)⁹]
= (7 x 8 + 9 x 8) + (21 x 8² - 36 x 8²) + &
= (56 + 72) + (21 – 36)8² + … = 128  + 64 (21 – 36) + …
= 64[2 + (21 – 36) + …]
Which is divisible by 64
Hence, the given statement is True.

Q.36. The number of terms in the expansion of [(2x + y³)⁴]⁷ is 8
Ans.
Given expression is [(2x + 3y)⁴]⁷ = (2x + 3y)²⁸
So, the number of terms = 28 + 1 = 29
Hence, the given statement is False.

Q.37. The sum of coefficients of the two middle terms in the expansion of (1 + x)^{2n – 1} is equal to ^{2n – 1}C_n.
Ans.
The given expression is (1 + x)^{2n – 1}
Number of terms = 2n – 1 + 1 = 2n (even)
∴ Middle terms areterm andterms
= nth terms and (n + 1)th terms
Coefficient of nth term = ^{2n – 1}C_n–1
and the coefficient of (n + 1)th term = ^{2n – 1}C_n
Sum of the coefficients = ^{2n – 1}C_{n – 1} + ^{2n – 1}C_n
= ^{2n – 1}C_n–1 + ^{2n – 1}C_n = ^{2n – 1 + 1}C_n = ²ⁿC_n
Hence, the statement [∴ ⁿC_r + ⁿC_{r -1} = ⁿ⁺¹C_r ] is False.  

Q.38. The last two digits of the numbers 3⁴⁰⁰ are 01. 
Ans.
Given    that    3⁴⁰⁰ = (9)²⁰⁰ = (10 – 1)²⁰⁰
∴ (10 – 1)²⁰⁰ = ²⁰⁰C₀(10)²⁰⁰ – ²⁰⁰C₁(10)¹⁹⁹ + … – ²⁰⁰C₁₉₉(10)¹ + ²⁰⁰C₂₀₀(1)²⁰⁰
= 10²⁰⁰ – 200 x 10¹⁹⁹ + & - 10 x 200 + 1
So, it is clear that last two digits are 01.
Hence, the given statement is True.

Q.39. If the expansion ofcontains a term independent of x, then n is a multiple of 2.
Ans.
The given expression is

For the term independent of x, 2n – 3r = 0
which not an integer and the expression is not possible to be true      Hence, the given statement is False.  

Q.40. Number of terms in the expansion of (a + b)ⁿ where n ∈ N is one less
than the power n.
Ans.
Since, the number of terms in the given expression (a + b)n is 1 more than n i.e., n + 1 Hence, the given statement is False.