Q.1. Find the term independent of x, x ≠ 0, in the expansion of.
Ans.
General Term Tr+1 = nCrxn–ryr
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 Tr+1 = nCr xn-r yr
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 Tr+1 = nCrxn–ryr
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 Tr+1 = nCrxn–r yr
Hence, the required middle term = – 252
(ii) Given expression is
Number of terms = 9 + 1 = 10 (even)
∴ Middle terms are
= 5th term and (5 + 1) = 6th term
General Term Tr+1 = nCrxn–ryr
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 Tr+1 = nCrxn–ryr
= 10Cr(x)10–r (–x2)r = 10Cr(x)10–r (–1)r × (x2)r
= (-1)r × 10Cr(x)10-r+2r = (- 1)r × 10Cr(x)10+r
To find the coefficient of x15, Put 10 + r = 15 ⇒ r = 5
∴ Coefficient of x15 = (-1)5 10C5 = - 10C5 = - 252
Hence, the required coefficient = - 252
Q.7. Find the coefficient ofin the expansion of
Ans.
The given expression is
General Term Tr+1 = nCrxn–ryr = 15Cr(4)15-r
= 15Cr (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 ( y1/2 + x1/3 )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 (y1/2 + x1/3)10
Sixth term is this expression
T6 = T5+1 = 10C5(y1/2)10–5 (x1/3)5 = 10C5y5/2 × x5/3
= 252 y5/2x5/3
Hence, the required term = 252 y5/2 . x5/3
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 Tr+1 = nCr xn–ryr
For coefficient of (2r + 4)th term, we have
T2r+4 = T2r+3+1 = 18C2r+3 (1)18–2r–3 × x2r+3
∴ Coefficient of (2r + 4)th term = 18C2r+3
Similarly, Tr-2 = Tr-3+1 = 18Cr-3 (1)18-r+3 × xr-3
∴ Coefficient of (r - 2)th term = 18Cr3
As per the condition of the questions, we have 18C2r+3 = 18Cr-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 + 2nC3
⇒
⇒
⇒ 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 Tr+1 = nCrxn–ryr
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 + nC1xn–1a + nC2xn–2a2 + nC3xn–3a3 + … + nCnan
Sum of odd terms,
O = nC0 xn + nC2 xn- 2 a2+ nC4 xn- 4 a4+ ...
and the sum of even terms
E = nC1 xn-1 × a + nC3 xn- 3a3+ nC5 xn- 5a5+ ...
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 xp occurs in the expansion of prove that its coefficient is
Ans.
Given expression is
General terms, Tr+1 = nCrxn–ryr
If xp occurs in
then 4n – 3r = p ⇒ 3r = 4n - p
⇒
∴ Coefficient of xp =
Hence, the coefficient of xp =
Q.17. Find the term independent of x in the expansion of (1 + x + 2x3)
Ans.
Given expression is
Let us consider
General Term Tr+1 = nCrxn–ryr
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 T3r = T3r–1+1 = 2nC3r–1 × x3r–1
and Tr+2 = Tr +1+1 = 2nCr +1 xr +1
As per the question, we have
As per the question, we have
2nC3r–1 = 2nCr+1
⇒ 3r – 1 + r + 1 = 2n [∴ nCp = nCq ⇒ 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 rth and (r + 1)th be two successive terms in the expansion (1 + x)24
∴ Tr+1 = 24Cr × xr
Tr+2 = Tr+1+1 = 24Cr+1xr+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 : 2nCn : 2n – 1Cn
Ans. (d)
Solution.
General Term Tr+1 = nCr xn–r yr
In the expansion of (1 + x)2n, we get Tr+1 = 2nCr xr
To get the coefficient of xn, put r = n
∴ Coefficient of xn = 2nCn
In the expansion of (1 + x)2n-1, we get Tr+1 = 2n-1Crxr
∴ Coefficient of xn is 2n-1Cn-1
The required ratio is
Hence, the correct option is (d).
Q.22. If the coefficients of 2nd, 3rd and the 4th 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 + … nCnxn 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
Tr+1 = 2nCrxr
∴ Coefficient of xn = 2nCn = A (Given)
In the expression (1 + x)2n–1
Tr+1 = 2n–1Crxr
∴ Coefficient of xn = 2n-1Cn = 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 = nCn/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 + … + nCn(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 Tr+1 be the constant term in the expansion of
For getting constant term, 32 – 4r = 0
⇒ = r = 8
∴ Tr + 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 + qCp = p + qCq]
Ans.
Given expansion is (1 + x)p+q
Tr+1 = p + qCr xr
Put r = p = p + qCpxp
∴ the coefficient of xp = p + qCp
Similarly, coefficient of xq = p + qCq
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 – 1Cn.
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 – 1Cn–1
and the coefficient of (n + 1)th term = 2n – 1Cn
Sum of the coefficients = 2n – 1Cn – 1 + 2n – 1Cn
= 2n – 1Cn–1 + 2n – 1Cn = 2n – 1 + 1Cn = 2nCn
Hence, the statement [∴ nCr + nCr -1 = n+1Cr ] 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.