Chapter 8 – Binomial Theorem

Class 11 Mathematics · 49 questions · 0 with answers

Solved examples

example-1Short answer

Find the r term in the expansion of x + . r −1 Solution We have Tr = 2rCr – 1 (x)2r – r + 1 2r = xr + 1 – r + 1 r −1 r +1 2r = x2 r −1 r +1

example-2Short answer

Expand the following (1 – x + x2)4 Solution Put 1 – x = y. Then (1 – x + x2)4 = (y + x2)4 = 4C0 y4 (x2)0 + 4C1 y3 (x2)1 + 4C2 y2 (x2)2 + 4C3 y (x2)3 + 4C4 (x2)4 = y4 + 4y3 x2 + 6y2 x4 + 4y x6 + x8 = (1 – x)4 + 4x2 (1 – x)3 + 6x4 (1 – x)2 + 4x6 (1 – x) + x8 = 1 – 4x + 10x2 – 16x3 + 19x4 – 16x5 + 10x6 – 4x7 + x8 x3 2

example-3Short answer

Find the 4th term from the end in the expansion of − 2 x2 Solution Since rth term from the end in the expansion of (a + b)n is (n – r + 2)th term from the beginning. Therefore 4th term from the end is 9 – 4 + 2, i.e., 7th term from the beginning, which is given by

example-4Short answer

Evaluate: x 2 − 1 − x 2 ) + (x + 1− x ) 2 2 BINOMIAL THEOREM 133 Solution Putting 1 − x 2 = y , we get The given expression = (x2 – y)4 + (x2 + y)4 = 2 [x8 + 4C2 x4 y2 + 4C4 y4] 8 4 ×3 4 = 2 x + x ⋅ (1 – x 2 ) + (1 − x 2 ) 2 2 ×1 = 2 [x8 + 6x4 (1 – x2) + (1 – 2x2 + x4] = 2x8 – 12x6 + 14x4 – 4x2 + 2

example-5Short answer

Find the coefficient of x11 in the expansion of x3 − x2 Solution Let the general term, i.e., (r + 1)th contain x11. 12 3 12 – r We have Tr + 1 = Cr (x ) − 2 = 12Cr x36 – 3r – 2r (–1)r 2r = 12Cr (–1)r 2r x36– 5r Now for this to contain x11, we observe that 36 – 5r = 11, i.e., r = 5 Thus, the coefficient of x11 is 12 × 11 × 10 × 9 × 8 C5 (–1)5 25 = − × 32 = –25344 5× 4 × 3× 2

example-6Short answer

Determine whether the expansion of x2 − will contain a term containing x10? Solution Let Tr + 1 contain x10. Then 18 2 18 − r −2 Tr + 1 = Cr ( x ) = 18Cr x36 – 2r (–1)r . 2r x– r = (–1)r 2r 18Cr x36 – 3r Thus, 36 – 3r = 10, i.e., r = Since r is a fraction, the given expansion cannot have a term containing x10. x 3

example-7Short answer

Find the term independent of x in the expansion of + 2 . 3 2x Solution Let (r + 1)th term be independent of x which is given by 10 − r r 10 x 3 Tr+1 = Cr 3 2 x2 10 − r x 2 1 = 10 Cr 32 3 2 x2r r 10 − r 10 − r − − 2r = 10 Cr 3 2 2 2− r x 2 Since the term is independent of x, we have

example-8Short answer

Find the middle term in the expansion of 2ax − . x2 Solution Since the power of binomial is even, it has one middle term which is the

example-9Short answer

Find the middle term (terms) in the expansion of + . Solution Since the power of binomial is odd. Therefore, we have two middle terms which are 5th and 6th terms. These are given by 5 4 9 p x p 126 p T5 = C4 = 9 C4 = x p x x 4 5 9 p x x 126 x and T6 = C5 = 9 C5 = x p p p

example-10Long answer

Show that 24n + 4 – 15n – 16, where n ∈ N is divisible by 225. Solution We have 24n + 4 – 15n – 16 = 24 (n + 1) – 15n – 16 = 16n + 1 – 15n – 16 = (1 + 15)n + 1 – 15n – 16 = n + 1C0 150 + n + 1C1 151 + n + 1C2 152 + n + 1C3 153 + ... + n + 1Cn + 1 (15)n + 1 – 15n – 16 = 1 + (n + 1) 15 + n + 1C2 152 + n + 1C3 153 + ... + n + 1Cn + 1 (15)n + 1 – 15n – 16 = 1 + 15n + 15 + n + 1C2 152 + n + 1C3 153 + ... + n + 1Cn + 1 (15)n + 1 – 15n – 16 = 152 [n + 1C2 + n + 1C3 15 + ... so on] Thus, 24n + 4 – 15n – 16 is divisible by 225.

example-11Long answer

Find numerically the greatest term in the expansion of (2 + 3x)9, where x= . 3x 9 Solution We have (2 + 3x) = 2 1 + 3x Tr + 1 29 9 Now, = r −1 Tr 3x 29 9 C r −1 9 r − 1 10 − r 3 x Cr 3x 9 = 9 = ⋅ Cr −1 2 r 9−r 9 2 10 − r 3 x 10 − r 9 3 = = Since x= r 2 r 4 2 Tr + 1 90 − 9r Therefore, ≥1 ⇒ ≥1 Tr 4r ⇒ 90 – 9r ≥ 4r (Why?) ⇒r≤ ⇒r≤6 Thus the maximum value of r is 6. Therefore, the greatest term is Tr + 1 = T7. 9 9 3x 3 Hence, T7 = 2 C6 , where x = 2 2 9 9 9 9 9 × 8 × 7 312 7 × 313 = 2 ⋅ C6 = 2 ⋅ = 4 3 × 2 × 1 212 2

example-12Long answer

If n is a positive integer, find the coefficient of x–1 in the expansion of (1 + x)n 1 + . Solution We have 1 n x +1 (1 + x) 2 n (1 + x) 1+ = (1 + x) = x x xn BINOMIAL THEOREM 137 –1 n Now to find the coefficient of x in (1 + x) 1+ , it is equivalent to finding (1 + x) 2 n coefficient of x–1 in which in turn is equal to the coefficient of xn – 1 in the expansion of (1 + x)2n. Since (1 + x)2n = 2nC0 x0 + 2nC1 x1 + 2nC2 x2 + ... + 2nCn – 1 xn–1 + ... + 2nC2n x2n Thus the coefficient of xn – 1 is 2nCn – 1 2n 2n = = n − 1 2n − n + 1 n −1 n +1

example-13Long answer

Which of the following is larger? 9950 + 10050 or 10150 We have (101)50 = (100 + 1)50 50.49 50.49.48 = 10050 + 50 (100)49 + (100)48 + (100) 47 + ... (1) 2.1 3.2.1 Similarly 9950 = (100 – 1)50 50.49 50.49.48 = 10050 – 50 . 10049 + (100)48– (100) 47 + ... (2) 2.1 3.2.1 Subtracting (2) from (1), we get  50 ⋅ 49 ⋅ 48 10150 – 9950 = 2 50 ⋅ (100) + 10047 + ...   3 ⋅ 2 ⋅1  50 ⋅ 49 ⋅ 48 ⇒ 10150 – 9950 = 10050 + 2 10047 + ... 3 ⋅ 2 ⋅1 ⇒ 10150 – 9950 > 10050 Hence 10150 >9950 + 10050

example-14Long answer

Find the coefficient of x50 after simplifying and collecting the like terms in the expansion of (1 + x)1000 + x (1 + x)999 + x2 (1 + x)998 + ... + x1000. Solution Since the above series is a geometric series with the common ratio , 1+ x its sum is 1001  1000 x   (1 + x) 1−   1+ x   x  1 −   1+ x  x 1001 (1 + x )1000 − 1+ x = = (1 + x)1001 – x1001 1+ x − x 1+ x Hence, coefficient of x50 is given by 1001 1001 C50 = 50 951

example-15Long answer

If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that a1 a3 2a2 + = a1 + a2 a3 + a4 a2 + a3 Solution Let a1, a2, a3 and a4 be the coefficient of four consecutive terms Tr + 1, Tr + , Tr + 3, and Tr + 4 respectively. Then a1 = coefficient of Tr + 1 = nCr a2 = coefficient of Tr + 2 = nCr + 1 a3 = coefficient of Tr + 3 = nCr + 2 and a4 = coefficient of Tr + 4 = nCr + 3 a1 Cr Thus = n a1 + a2 Cr + n Cr + 1 = n +1 (∵ n Cr + nCr + 1 = n + 1Cr + 1 ) Cr + 1 BINOMIAL THEOREM 139 n r +1 n − r r +1 = × = r n−r n +1 n +1 a3 Cr + 2 Similarly, = n a3 + a4 Cr + 2 + n Cr + 3 Cr + 2 r+3 = n +1 = Cr + 3 n +1 a1 a3 r + 1 r + 3 2r + 4 Hence, L.H.S. = + = + = a1 + a2 a3 + a4 n + 1 n + 1 n + 1 2a2 2 n Cr + 1 ( 2 n Cr + 1 ) ( ) and R.H.S. = =n = n +1 a2 + a3 Cr + 1 + n Cr + 2 Cr + 2 n r + 2 n − r − 1 2 (r + 2) = 2 × = r +1 n − r −1 n +1 n +1

example-16Multiple choice

The total number of terms in the expansion of (x + a)51 – (x – a)51 after simplification is

  • (a)102
  • (b)25
  • (c)26
  • (d)None of these Solution C is the correct choice since the total number of terms are 52 of which 26 terms get cancelled.
example-17Multiple choice

If the coefficients of x and x in 2 + 7 8 are equal, then n is

  • (a)56
  • (b)55
  • (c)45
  • (d)15 n–r r Solution B is the correct choice. Since Tr + 1 = Cr a x in expansion of (a + x)n, n n−7 x 2n − 7 7 Therefore, T8 = C7 (2) = n C7 x 3 37 x n 2n − 8 8 and T9 = n C8 (2) n–8 = C8 8 x 3 3 n 2n − 7 n 2n − 8 Therefore, C7 7 = C8 8 (since it is given that coefficient of x7 = coefficient x8) 3 3 n 8 n − 8 2n − 8 37 ⇒ × = ⋅ 7 n−7 n 38 2n − 7 8 1 ⇒ = ⇒ n = 55 n−7 6
example-18Multiple choice

If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n, then a0 + a2 + a4 + ... + a2n equals. 3n + 1 3n − 1 1 − 3n n 1

  • (A)
  • (B)
  • (C)
  • (D)3 + 2 2 2 2 Solution A is the correct choice. Putting x = 1 and –1 in (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n we get 1 = a0 + a1 + a2 + a3 + ... + a2n ... (1) and 3n = a0 – a1 + a2 – a3 + ... + a2n ... (2) Adding (1) and (2), we get 3n + 1 = 2 (a0 + a2 + a4 + ... + a2n) 3n + 1 Therefore a0 + a2 + a4 + ... + a2n =
example-19Multiple choice

The coefficient of xp and xq (p and q are positive integers) in the expansion of (1 + x)p + q are

  • (A)equal
  • (B)equal with opposite signs
  • (C)reciprocal of each other
  • (D)none of these Solution A is the correct choice. Coefficient of xp and xq in the expansion of (1 + x)p +q are p + qCp and p + qCq p+q p+q and Cp = p + qCq = Hence (a) is the correct answer. BINOMIAL THEOREM 141
example-20Multiple choice

The number of terms in the expansion of (a + b + c)n, where n ∈ N is (n + 1) (n + 2)

  • (A)
  • (B)n + 1
  • (C)n + 2
  • (D)(n + 1) n Solution A is the correct choice. We have (a + b + c)n = [a + (b + c)]n = an + nC1 an – 1 (b + c)1 + nC2 an – 2 (b + c)2 + ... + nCn (b + c)n Further, expanding each term of R.H.S., we note that First term consist of 1 term. Second term on simplification gives 2 terms. Third term on expansion gives 3 terms. Similarly, fourth term on expansion gives 4 terms and so on. The total number of terms = 1 + 2 + 3 + ... + (n + 1) (n + 1) (n + 2) =
example-21Multiple choice

The ratio of the coefficient of x15 to the term independent of x in 2 2 x + is

  • (A)12:32
  • (B)1:32
  • (C)32:12
  • (D)32:1 2 2 Solution (B) is the correct choice. Let Tr + 1 be the general term of x + , so, 15 2 15 – r 2 Tr + 1 = Cr (x ) = 15Cr (2)r x30 – 3r ... (1) Now, for the coefficient of term containing x15,
example-22Multiple choice

If z = 3 i 3 i , then + + − 2 2 2 2

  • (A)Re (z) = 0
  • (B)Im (z) = 0
  • (C)Re (z) > 0, Im (z) > 0
  • (D)Re (z) > 0, Im (z) < 0 Solution B is the correct choice. On simplification, we get 2 3  5  3  5  3  i 5  3 i 4  z = 2  C0   + C2   + C4     2 2 2  2 2  Since i2 = –1 and i4 = 1, z will not contain any i and hence Im (z) = 0.

Questions

Q8.1Long answer

Overview: 8.1.1 An expression consisting of two terms, connected by + or – sign is called a 1 1 4 binomial expression. For example, x + a, 2x – 3y, − 3 , 7x − , etc., are all binomial x x 5y expressions. 8.1.2 Binomial theorem If a and b are real numbers and n is a positive integer, then (a + b)n =nC0 an + nC1 an – 1 b1 + nC2 an – 2 b2 + ... ... + nCr an – r br + ... + nCn bn, where nCr = for 0 ≤ r ≤ n r n−r The general term or (r + 1)th term in the expansion is given by Tr + 1 = nCr an–r br 8.1.3 Some important observations 1. The total number of terms in the binomial expansion of (a + b)n is n + 1, i.e. one more than the exponent n.

Q2Long answer

In the expansion, the first term is raised to the power of the binomial and in each subsequent terms the power of a reduces by one with simultaneous increase in the power of b by one, till power of b becomes equal to the power of binomial, i.e., the power of a is n in the first term, (n – 1) in the second term and so on ending with zero in the last term. At the same time power of b is 0 in the first term, 1 in the second term and 2 in the third term and so on, ending with n in the last term.

Q3Short answer

In any term the sum of the indices (exponents) of ‘a’ and ‘b’ is equal to n (i.e., the power of the binomial).

Q4Short answer

The coefficients in the expansion follow a certain pattern known as pascal’s triangle. Index of Binomial Coefficient of various terms 0 1 1 1 1 2 1 2 1 3 1 3 3 1 4 1 4 6 4 1

Q5Multiple choice

1 5 10 10 5 1 Each coefficient of any row is obtained by adding two coefficients in the preceding row, one on the immediate left and the other on the immediate right and each row is bounded by 1 on both sides. The (r + 1)th term or general term is given by Tr + 1 = n Cr an – r br 8.1.4 Some particular cases If n is a positive integer, then (a + b)n = nC0 an b0 + nC1 an b1 + nC2 an – 2 b2 + ... + nCr an – r br + ... + Cn a0 bn ... (1) In particular 1. Replacing b by – b in (i), we get (a – b)n = nC0 an b0 – nC1 an – 1 b1 + nC2 an – 2 b2 + ... + (–1)r nCr an – r br + ... + (–1)n nCn a0 bn ... (2) 2. Adding (1) and (2), we get (a + b)n + (a – b)n = 2 [nC0 an b0 + nC2 an – 2 b2 + nC4 an – 4 b4 + ... ] = 2 [terms at odd places] 3. Subtracting (2) from (1), we get (a + b)n – (a – b)n = 2 [nC1 an – 1 b1 + nC3 an – 3 b3 + ... ] = 2 [sum of terms at even places] 4. Replacing a by 1 and b by x in (1), we get (1 + x)n =nC0 x0 + nC1 x + nC2 x2 + ... + nCr xr + ... + nCn – 1 xn – 1 + nCn xn i.e. (1 + x)n = Cr x r r =0 BINOMIAL THEOREM 131 5. Replacing a by 1 and b by –x in ... (1), we get (1 – x)n = nC0 x0 – nC1 x + nC2 x2 ... + nCn–1 (–1)n–1 xn-1 + nCn (–1)n xn i.e., (1 – x)n = (−1) r n Cr x r r =0 8.1.5 The pth term from the end The pth term from the end in the expansion of (a + b) n is (n – p + 2)th term from the beginning. 8.1.6 Middle terms The middle term depends upon the value of n.

  • (a)If n is even: then the total number of terms in the expansion of (a + b)n is n + 1 (odd). Hence, there is only one middle term, i.e., n term is the middle +1 term.
  • (b)If n is odd: then the total number of terms in the expansion of (a + b)n is n + 1 th th n +1 n+3 (even). So there are two middle terms i.e., and are two 2 2 middle terms. 8.1.7 Binomial coefficient In the Binomial expression, we have (a + b)n =nC0 an + nC1 an – 1 b + nC2 an – 2 b2 + ... + nCn bn ... (1) n n n n The coefficients C0, C1, C2, ... , Cn are known as binomial or combinatorial coefficients. Putting a = b = 1 in (1), we get C0 + nC1 + nC2 + ... + nCn = 2n Thus the sum of all the binomial coefficients is equal to 2n. Again, putting a = 1 and b = –1 in (i), we get C0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even 2n binomial coefficients and each is equal to = 2 n−1 . C0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... = 2n – 1
Q9Short answer

x3 −2 9 x 9 64 9 × 8 × 7 64 672 T7 = C6 = C3 ⋅ = × = 3 2 x2 8 x12 3 × 2 × 1 x 3 x 4 4 (

Q10Short answer

− r − 2r = 0 ⇒ r= 2 Hence 3rd term is independent of x and its value is given by 10 3−3 10 × 9 1 5 T3 = C2 = × = 4 2 × 1 9 × 12 12

Q12Short answer

+ 2 term and it is given by 12 −b T7 = C6 (2ax)6 x2 12 26 a 6 x 6 ⋅ (−b)6 = C6 x12 12 26 a 6b6 59136a 6b6 = C6 = x6 x6 BINOMIAL THEOREM 135

Q30Multiple choice

– 3r = 15, i.e., r = 5 Therefore, C5 (2) is the coefficient of x15 (from (1)) 15 5 To find the term independent of x, put 30 – 3r = 0 Thus 15C10 210 is the term independent of x (from (1)) C5 25 1 1 Now the ratio is 15 10 = 5= C10 2 2 32 5 5

Q1Short answer

Find the term independent of x, x ≠ 0, in the expansion of − . 2 3x

Q2Short answer

If the term free from x in the expansion of x− is 405, find the value x2 of k.

Q3Short answer

Find the coefficient of x in the expansion of (1 – 3x + 7x2) (1 – x)16.

Q4Short answer

Find the term independent of x in the expansion of, 3x − . x2

Q5Multiple choice

Find the middle term (terms) in the expansion of 10 9 x a  x3 

  • (i) − 
  • (ii)3x −  a x 6 BINOMIAL THEOREM 143
Q6Short answer

Find the coefficient of x15 in the expansion of (x – x2)10. 1 1

Q7Short answer

Find the coefficient of in the expansion of x4 − . x17 x3 1 1 n (

Q8Short answer

Find the sixth term of the expansion y 2 + x 3 ) , if 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.]

Q9Short answer

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.

Q10Short answer

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.

Q11Short answer

Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.

Q12Long answer

If p is a real number and if the middle term in the expansion of +2 is 1120, find p. 2n

Q13Long answer

Show that the middle term in the expansion of x− is 1 × 3 × 5 × ... (2n − 1) × ( −2) n . 3 1

Q14Long answer

Find n in the binomial 2+ 3 if the ratio of 7th term from the beginning to the 7th term from the end is .

Q15Multiple choice

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 2n 2 1
Q16Long answer

If x occurs in the expansion of x + , prove that its coefficient is 2n . 4n − p 2n + p 3 3 3 1 

Q17Long answer

Find the term independent of x in the expansion of (1 + x + 2x )  x 2 −  . 2 3x 

Q18Multiple choice

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
Q19Multiple choice

Given the integers r > 1, n > 2, and coefficients of (3r) 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
Q20Multiple choice

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 Cr 1 r +1 1 [Hint: 24 = ⇒ 4r + 4 = 24 – 4 ⇒ r = 4 ] Cr + 1 4 24 − r 4
Q21Multiple choice

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 2n 2n – 1 [Hint : Cn : Cn ]
Q22Multiple choice

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 n n n 2 [Hint: 2 C2 = C1 + C3 ⇒ n – 9n + 14 = 0 ⇒ n = 2 or 7] BINOMIAL THEOREM 145
Q23Multiple choice

If A and B are coefficient of xn in the expansions of (1 + x)2n and (1 + x)2n – 1 respectively, then equals 1 1

  • (A)1
  • (B)2
  • (C)
  • (D)2 n 2n A C [Hint: = 2n −1 n = 2 ] B Cn 1 7
Q24Multiple choice

If the middle term of + x sin x is equal to 7 , then value of x is x 8 π π π π

  • (A)2nπ +
  • (B)nπ +
  • (C)nπ + (–1)n
  • (D)nπ + (–1)n 6 6 6 3 1 5 63 1 1 [Hint: T6 = 10 C5 ⋅ x sin 5 x = ⇒ sin5 x = 5 ⇒ sin x = x 5 8 2 2 π ⇒ x = nπ + (–1)n ] Fill in the blanks in Exercises 25 to 33.
Q25Fill in the blanks

The largest coefficient in the expansion of (1 + x)30 is _________________ .

Q26Fill in the blanks

The number of terms in the expansion of (x + y + z)n _________________ . [Hint: (x + y + z)n = [x + (y + z)]n]

Q27Fill in the blanks

In the expansion of x2 − , the value of constant term is x2 _________________ .

Q28Fill in the blanks

If the seventh terms from the beginning and the end in the expansion of 3 1 2+ 3 are equal, then n equals _________________ . 1 n−6 6 1 6 n−6 [Hint : T7 = T n – 7 + 2 ⇒ C ( ) 6 23 = C (2 ) 1 n −6 33 33 n − 12 1 n − 12 1 ⇒ ( ) 23 = 1 ⇒ only problem when n – 12 = 0 ⇒ n = 12]. –6 4 1 2b

Q29Fill in the blanks

The coefficient of a b in the expansion of − is _________. a 3 b 4 10 1  −2b 1120 − 6 4 [Hint : T5 = C 4  3  = 27 a b ]

Q30Fill in the blanks

Middle term in the expansion of (a3 + ba)28 is _________ .

Q31Fill in the blanks

The ratio of the coefficients of xp and xq in the expansion of (1 + x)p + q is_________ [Hint: p + qCp = p + qCq] x 3

Q32Fill in the blanks

The position of the term independent of x in the expansion of + 2 is 3 2x _________ .

Q33Fill in the blanks

If 2515 is divided by 13, the reminder is _________ . State which of the statement in Exercises 34 to 40 is True or False. 10 20 20 C10

Q34Multiple choice

The sum of the series ∑ C r is 219 + r =0 2

Q35Multiple choice

The expression 79 + 97 is divisible by 64. Hint: 79 + 97 = (1 + 8)7 – (1 – 8)9

Q36Multiple choice

The number of terms in the expansion of [(2x + y3)4]7 is 8

Q37Multiple choice

The sum of coefficients of the two middle terms in the expansion of (1 + x)2n – 1 is equal to 2n – 1Cn.

Q38Multiple choice

The last two digits of the numbers 3400 are 01. 2n

Q39Multiple choice

If the expansion of x − 2 contains a term independent of x, then n is a multiple of 2.

Q40Multiple choice

Number of terms in the expansion of (a + b)n where n ∈ N is one less than the power n.