• Chapter 8.1 Overview: 8.1.1 An expression consisting of two terms, connected by + or – sign is called a 1 1 4binomial expression. For example, x+a, 2x– 3y, − 3,7x− , etc., are all binomial x x 5y expressions. 8.1.2 Binomial theorem If aand bare real numbers and n is a positive integer, then n– 1 bn– 2 b2(a+ b)n =nC0 an+ nC1 a1 + nC2 a + ... n– n... + nC arbr+... + nC b, where nC = n for 0 ≤ r≤ n r nrrn r−The general term or (r+ 1)th term in the expansion is given by nn–rbrT = Car+ 1r8.1.3 Some important observations 1. The total number of terms in the binomial expansion of (a+ b)nis n+ 1, i.e. one more than the exponent n. 2. In the expansion, the first term is raised to the power of the binomial and in each subsequent terms the power of areduces by one with simultaneous increase in the power of bby one, till power of bbecomes equal to the power of binomial, i.e., the power of ais nin 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 bis 0 in the first term, 1 in the second term and 2 in the third term and so on, ending with nin the last term. 3. In any term the sum of the indices (exponents) of ‘a’ and ‘b’ is equal to n(i.e., the power of the binomial). 4. 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 4 1 1 4 3 3 1 6 4 1 5 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 nC an – r brT = r + 1r 8.1.4 Some particular cases If n is a positive integer, then an b0an an – 2 nC an – r(a + b)n = nC + nCb1 + nCb2 + ... + br + ... +0 12 r a0 bnnCn ... (1) In particular 1. Replacing b by – b in (i), we get (a – b)n = nCan b0 – nCan – 1 b1 + nCan – 2 b2 + ... + (–1)rn C an – r br + ... +01 2 r(–1)nnCna0 bn ... (2) 2. Adding (1) and (2), we get an – 2 b2 + nC(a + b)n + (a – b)n =2 [nCan b0 + nCan – 4 b4 + ... ]02 4 = 2 [terms at odd places] 3. Subtracting (2) from (1), we get an – 1 b1 + n Can – 3 b3(a + b)n – (a – b)n =2 [nC + ... ]13 = 2 [sum of terms at even places] 4. Replacing a by 1 and b by x in (1), we get 1 xn – 1(1 + x)n =nC0 x0 + nC1 x + nC2 x2 + ... + nCxr + ... + nC + nCxn rn – n n i.e. (1 + x)n =∑ 0=r n C xr r BINOMIAL THEOREM 131 5. Replacing a by 1 and b by –x in ... (1), we get (1 – x)n = nCx0 – nCx + nCx2 ... + n C (–1)n–1 xn-1 + nC (–1)n xn 0 12 n–1 nn i.e., (1 – x)n = ∑(1) − rnCr xr 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 ⎛ n ⎞th (odd). Hence, there is only one middle term, i.e., ⎜+ 1⎟ term is the middle ⎝ 2 ⎠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 =n Can + nCan – 1 b + nCan – 2 b2 + ... + nC bn ... (1)01 2 n The coefficients nC0, nC1, nC2, ... , n C n are known as binomial or combinatorial coefficients. Putting a = b = 1 in (1), we get nC0 + nC1 + nC2 + ... + n Cn = 2n Thus the sum of all the binomial coefficients is equal to 2n . Again, putting a = 1 and b = –1 in (i), we get nC + nC + nC + ... = nC + nC + nC+ ...024135 Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even 2nn−1binomial coefficients and each is equal to = 2.2 nC0 + nC2 + nC4 + ... = nC1 + nC3 + nC5 + ... = 2n – 1 8.2 Solved Examples Short Answer Type ⎛ 1 ⎞2r Example 1 Find the rth term in the expansion of ⎜x +⎟.⎝ x ⎠r −11⎛⎞(x)2 r – r + 1Solution We have T =2rC ⎜⎟rr – 1x⎝⎠ 2r xr + 1 – r + 1= r −1 r +1 2r 2 = x r −1 r +1 Example 2 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 3⎛x 2 ⎞9 Example 3 Find the 4th term from the end in the expansion of ⎜−2 ⎟⎝2 x ⎠ 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 36 9 x 32 x 64 9 ××87 64⎛⎞ −672 9 ⎛⎞9CC ⋅= ×=T = 6 ⎜⎟⎜ ⎟= 312 3372⎝⎠⎝ ⎠2 x 8 x 3 ××12 xx 222 2Example 4 Evaluate: (x 1 x )4 +(x 1−− +− x )4 BINOMIAL THEOREM 133 Solution Putting 1 −x2 =y , we get The given expression = (x2 – y)4 + (x2 + y)4 =2 [x8 + 4C2 x4 y2 + 4C4 y4] ⎡×8434 2 22 ⎤ =2 x + x ⋅(1 – x ) +− x )(1 ⎢⎥⎣ 21×⎦ =2 [x8 + 6x4 (1 – x2) + (1 – 2x2 + x4] =2x8 – 12x6 + 14x4 – 4x2 + 2 ⎛2 12 Example 5 Find the coefficient of x11 in the expansion of ⎜⎝x3 −2 ⎞⎟ x ⎠ Solution Let the general term, i.e., (r + 1)th contain x11. r = 12C (x3)12 – r ⎛ 2 ⎞We have Tr + 1r⎜−2 ⎟⎝ x ⎠ = 12Cx36 – 3r – 2rr (–1)r 2r = 12Cx36– 5r r (–1)r 2r 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 12C5 (–1)5 25 = −×32 = –253445432×××Example 6 Determine whether the expansion of ⎛⎜x2 −2⎞18 will contain a term⎟⎝⎠xcontaining x10? Solution Let T contain x10. Thenr + 1⎛⎞18 218 −r −2 r T =C(x ) ⎜⎟rr + 1⎝⎠ x = 18C x36 – 2r r r (–1)r . 2r x–18Cx36 – 3r= (–1)r 2rr 26Thus, 36 – 3r = 10, i.e., r = 3 Since ris a fraction, the given expansion cannot have a term containing x10. 10⎛x 3 ⎞Example 7 Find the term independent of xin the expansion of ⎜+ .2 ⎟ ⎝32x ⎠ Solution Let (r+ 1)th term be independent of xwhich is given by 10 −r 10 ⎛ x⎞⎛ 3 ⎞r T r+1 = Cr ⎜⎟ ⎜ 2 ⎟⎝ 3 ⎠⎝2x ⎠ 10 −r = 10 ⎛⎞ 2x 2 r⎛ 1 ⎞Cr ⎜⎟ 3 ⎜r ⎟⎝⎠ 3 ⎝2 x2r⎠ r 10 −r 10 −r−−2r =10 22−r 2C3 2 xr Since the term is independent of x, we have 10 −r−2r=0 ⇒ r=22 Hence 3rd term is independent of xand its value is given by 10 3−3 10 ×91 5C =×=T3 = 2 42 ×19 ×12 12 ⎛−b⎞12 Example 8 Find the middle term in the expansion of ⎜2ax 2 ⎟ .⎝ x ⎠ Solution Since the power of binomial is even, it has one middle term which is the ⎛12 +2 ⎞th term and it is given by⎜⎟⎝ 2 ⎠⎛⎞12 6 −b 6 T7 = C(2 ax)26 ⎜⎟x⎝⎠ 666 6 12 2 ax ⋅−)( b = C6 12x 666 66 12 2 ab 59136 ab =C = 66 6xx BINOMIAL THEOREM 135 ⎛px ⎞9 Example 9 Find the middle term (terms) in the expansion of ⎜+⎟. ⎝xp ⎠ Solution Since the power of binomial is odd. Therefore, we have two middle terms which are 5th and 6th terms. These are given by 9 p 5 x 49 p 126 ⎛⎞⎛⎞ pT =C ⎜⎟=C = 54 ⎜⎟ 4xp x⎝⎠⎝⎠ x p 4 x 5 x 126 x9 ⎛⎞⎛⎞ 9and T = C5 =C5 =⎜⎟⎝⎠⎜⎟⎝⎠6xp pp Example 10 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 + 1C 150 + n + 1C 151 + n + 1C 152 + n + 1C=0123 153 + ... + n + 1C (15)n + 1 – 15n – 16n + 1 152 + n + 1C3=1 + (n + 1) 15 + n + 1C2 153 + ... + n + 1C (15)n + 1 – 15n – 16 n + 1 152 + n + 1C=1 + 15n + 15 + n + 1C23 153 + ... + n + 1C (15)n + 1 – 15n – 16 n + 1=152 [n + 1C2 + n + 1C3 15 + ... so on] Thus, 24n + 4 – 15n – 16 is divisible by 225. Long Answer Type Example 11 Find numerically the greatest term in the expansion of (2 + 3x)9, where 3 x = .2 9 ⎛ 3x ⎞9 21Solution We have (2 + 3x)9 = ⎜+ ⎟⎝ 2 ⎠ ⎡ r⎤99 3x⎛⎞ 2 ⎢Cr⎜ ⎟⎥Tr+1 ⎣ ⎝⎠ ⎦2Now, = r−1T ⎡r 99 ⎛⎞ 3x ⎤ 2 ⎢C ⎜ ⎟ ⎥r−1⎣ ⎝⎠ 2 ⎦ 9C 3xr = 29Cr−1 10 −rx3 = r 2 T 90 −9r r−110 −r9 3x⋅= r9 −r 9 2 10 r93−⎛⎞ = ⎜⎟Since x= r 42⎝⎠ r+1 ≥ ⇒Therefore, 1 ≥1Tr 4r ⇒ 90 – 9r≥ 4r (Why?) 90 ⇒ r≤ 13 12 ⇒ r≤ 6 13 Thus the maximum value of ris 6. Therefore, the greatest term is T = T. r+ 17⎡ 6 99 3x 3⎛ ⎞⎤ Hence, T =2 ⎢C ⎜ ⎟⎥, where x= 7⎣ 62 2⎝ ⎠ ⎦6 12 13 9 9873××⎛ ⎞ 7399 ⎛⎞ 9 × =2 ⋅C6 ⎜⎟= 2 ⋅⎜12 ⎟= 4 3212 ⎠ 2⎝⎠××⎝ Example 12 If n is a positive integer, find the coefficient of x–1 in the expansion of ⎛ 1 ⎞n (1 + x)n⎜1 +⎟.⎝ x⎠Solution We have nn 2n⎛ 1⎞⎛x+1⎞ (1 +x)(1 + x)n⎜1+⎠⎟ = (1 + x)n⎜⎟= n⎝ x ⎝x⎠x BINOMIAL THEOREM 137 ⎛ 1⎞n Now to find the coefficient of x–1 in (1 + x)n 1 , it is equivalent to finding⎜+⎟⎝ x⎠(1+x)2 n coefficient of x–1 in which in turn is equal to the coefficient of xn – 1 in thenxexpansion of (1 + x)2n . Since (1 + x)2n = 2nC0 x0 + 2n C1 x1 + 2nC2 x2 + ... + 2nCn – 1 xn –1 + ... + 2nC2 nx2 n Thus the coefficient of xn – 1 is 2nC n – 1 2n2n = = n −1 n +1n −12nn 1−+Example 13 Which of the following is larger? 9950 + 10050 or 10150 We have (101)50 = (100 + 1)50 50.49 50.49.48 47 = 10050 + 50 (100)49 + (100)48 + (100) + ... (1)2.13.2.1 Similarly 9950 = (100 – 1)50 50.49 50.49.48 = 10050 – 50 . 10049 + (100)48– (100) 47 + ... (2) 2.13.2.1 Subtracting (2) from (1), we get 5049 48 ⎤⎡ ⋅⋅ 49 472 50 (100) + ...10150 – 9950 = ⋅ 100 +⎢⎥321 ⎣⎦ ⎛50 49 ⋅⋅ 48⎞ 47⇒ 10150 – 9950 = 10050 + 2 100 +... ⎜⎝ 321 ⋅⋅ ⎟⎠ ⇒ 10150 – 9950 > 10050 Hence 10150 >9950 + 10050 Example 14 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. x Solution Since the above series is a geometric series with the common ratio 1 +,x its sum is 1000 ⎡⎛ x ⎞1001 ⎤ (1+ x) ⎢1 −⎥⎜⎟⎝1 + x ⎠⎣⎦ ⎡⎛ x ⎞⎤1⎢ −⎜ ⎟⎥⎝1 + x⎠⎣⎦1001 x10 00 (1 + x) − 1 + x x1001= = (1 + x)1001 – 1 + x − x 1 + xHence, coefficient of x50 is given by 1001 1001C50 = 50 951 Example 15 If a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that a1 a32a2+= a + aa + aa + a12 34 23 Solution Let a1, a2, a3 and a4 be the coefficient of four consecutive terms T, Tr + 1r + , T, and T respectively. Then 2r + 3r + 4a= coefficient of T = nC1 r + 1r a= coefficient of T = nC2 r + 2r + 1 a= coefficient of T = nC3 r + 3r + 2 and a4 = coefficient of Tr + 4 = nCr + 3 a1 n CrThus = a + an n 1 2Cr + Cr + 1 n Cr n C = n + 1C= (∵ + nC)n +1 rr + 1r + 1Cr +1 BINOMIAL THEOREM 139 Similarly, 3 3 4 a a a+ = = = n rn r− × 1 1 r n r n + − +2 2 3 C C C n r n n r r + + ++ 2 1 3 C C n r n r + + + = 3 1 r n + + = 1 1 r n ⎛+⎞ ⎜ ⎟⎝+⎠ a ar+1 r+32r+4 Hence, L.H.S. = 1 + 3 =+= a+aa+an+1 n+1 n+112 34 nn2C 22a2 (r+1 )(Cr+1 )and R.H.S. = == nn n+1a+a C +CC23 r+1 r+2 r+2 r+2 −− n nr 1 2(r+2) =2 × = r+1 nr 1 n+1 n+1−− Objective Type Questions (M.C.Q) Example 16 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. ⎛ x⎞n Example 17 If the coefficients of x7 and x8 in ⎜2 +⎠⎟are equal, then nis⎝ 3 (a) 56 (b) 55 (c) 45 (d) 15 n– rxSolution B is the correct choice. Since T = nCarin expansion of (a+ x)n ,r+ 1rn−7x 2nn−7 ⎛⎞7 n 7Therefore, = C (2) =C xT87 ⎜⎟7733⎝⎠ n −8x 82⎛⎞n 8and T9 = nC8 (2)n–8 ⎜⎟=C8 x 3 38⎝⎠ n −7 n −8 n 2 n 2Therefore, C7 3 = C8 3 (since it is given that coefficient ofx7 = coefficientx8)78 n −8 n −878n 23 ⇒ × = ⋅8 n −77 n −7 n 32 81 ⇒ = ⇒ n = 55 n −76Example 18 If (1 – x + x2)n = a + ax + ax2 + ... + a x2n, then a + a + a + ...012 2n024+ a equals.2nnn n3 +13 −1 −113 n(A) (B) (C) (D) 3 + 222 2 Solution A is the correct choice. Putting x = 1 and –1 in (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2 nx2 n 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(a + a + a + ... + a)0242 n3n +1 Therefore a0 + a2 + a4 + ... + a2n = 2 Example 19 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 + q C and p + qC pq p +q p + q C = p + qCand pq = p q Hence (a) is the correct answer. BINOMIAL THEOREM 141 Example 20 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 2 (a + b + c)n =[a + (b + c)]n = an + nC1 an – 1 (b + c)1 + n C2 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) = 2 Example 21 The ratio of the coefficient of x15 to the term independent of x in ⎛22⎞15 ⎜x +⎟ is⎝ x⎠(A) 12:32 (B) 1:32 (C) 32:12 (D) 32:1 ⎛22⎞15 Solution (B) is the correct choice. Let Tr + 1 be the general term of ⎜x +⎟, so,⎝ x⎠2 r⎛⎞(x2)15 – rT = 15C r + 1r⎜⎟⎝⎠x = 15C r (2)rx30 – 3r ... (1) Now, for the coefficient of term containing x15, 30 – 3r = 15, i.e., r = 5 Therefore, 15C5 (2)5 is the coefficient of x15 (from (1)) To find the term independent of x, put 30 – 3r = 0 Thus 15C10 210 is the term independent of x (from (1)) 15C255 1 1 Now the ratio is == 15 105C2 23210 ⎛3 i ⎞5 ⎛3 i ⎞5 Example 22 If z = −⎟, then⎝22⎠⎝22⎠+⎟+⎜⎜ (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 ⎡ ⎛3 ⎞2 ⎛3⎞3 ⎛⎞i 2 ⎛3 ⎞⎛⎞i 4 ⎤ ⎢5z = 2C0 ⎜ ⎟+5C2 ⎜ ⎟⎜⎟ +5C4 ⎜ ⎟⎜⎟ ⎥⎝2 ⎠ ⎝ ⎠⎝⎠ 22 ⎝⎠ ⎣ 2 ⎝⎠2 ⎦ Since i2 = –1 and i4 = 1, z will not contain any i and hence I (z) = 0.mShort Answer Type ⎛3x21 ⎞15 1. Find the term independent of x, x ≠ 0, in the expansion of ⎜ −⎟.⎝23x ⎠k ⎞10 2. If the term free from x in the expansion of ⎜⎛ x −2 ⎟⎠ is 405, find the value⎝ x of k. 3. Find the coefficient of x in the expansion of (1 – 3x + 7x2) (1 – x)16. ⎛ 2 ⎞15 4. Find the term independent of x in the expansion of, ⎜3x −⎟ x2 ⎠ .⎝ 5. Find the middle term (terms) in the expansion of ⎛xa⎞10 ⎛ x3 ⎞9 (i) ⎜−⎟ (ii) ⎜3x −⎟⎝ax ⎠⎝ 6 ⎠ BINOMIAL THEOREM 143 6. Find the coefficient of x15 in the expansion of (x – x2)10. 1 15⎛41 ⎞7. Find the coefficient of 17 in the expansion of ⎜x − 3 ⎟⎠ . x ⎝ x n1 1 8. Find the sixth term of the expansion (), if the binomial coefficient ofy 2 +x3 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.] 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. 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. 11. Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11. Long AnswerType ⎛p ⎞8 12. If p is a real number and if the middle term in the expansion of ⎜+2⎟⎠ is⎝2 1120, find p. ⎛ 1 ⎞2n 13. Show that the middle term in the expansion of ⎜x −⎟ is ⎝ x ⎠ 13××× 5... (2 n −1) (2)n×− .n 1 ⎞n⎛ 14. Find n in the binomial ⎜32 +3 ⎟⎠ if the ratio of 7th term from the beginning to⎝ 3 1 the 7th term from the end is 6. 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)2 n ⎛21⎞2n 16. If xp occurs in the expansion of ⎜x +⎟⎠, prove that its coefficient is⎝ x 2n .4n − p 2n +p 3 3 ⎛31 ⎞9 17. Find the term independent of x in the expansion of (1 + x + 2x3) ⎜x2 −⎟.⎝23x⎠Objective Type Questions Choose the correct answer from the given options in each of the Exercises 18 to 24 (M.C.Q.). a)10018. The total number of terms in the expansion of (x + a)100 + (x – after simplification is (A) 50 (B) 202 (C) 51 (D) none of these 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 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 24Cr 1 r +11[Hint: =⇒ = ⇒ 4r + 4 = 24 – 4 ⇒ ]r =424C 424 −r 4r +1 21. The coefficient of xn in the expansion of (1 + x)2 n and (1 + x)2 n – 1 are in the ratio. (A) 1 : 2 (B) 1 : 3 (C) 3 : 1 (D) 2 : 1 : 2 n – 1C[Hint : 2n Cnn 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: 2 nC2 = nC1 + n C3 ⇒ n2 – 9n + 14 = 0 ⇒ n = 2 or 7 BINOMIAL THEOREM 145 23. If A and B are coefficient of xn in the expansions of (1 + x)2n and (1 + x)2n – 1 A respectively, then B equals 1 1 (A) 1 (B)2 (C) (D)2 n A2n Cn[Hint: = 2n −1 = 2]BCn 10 7⎛ 1 ⎞24. If the middle term of ⎜+ x sin x⎟ is equal to 7 8 , then value of x is⎝ x ⎠ππ ππ (A) 2nπ + (B) nπ + (C) nπ + (–1)n (D) nπ + (–1)n 66 63 11 [Hint: T6 = 10C5 15 ⋅ x5 sin 5 x = 63 ⇒ sin5 x = 5 ⇒ sin = x 8 22 π ⇒ x = nπ + (–1)n ]6 Fill in the blanks in Exercises 25 to 33. 25. The largest coefficient in the expansion of (1 + x)30 is _________________ . 26. The number of terms in the expansion of (x + y + z)n _________________ . [Hint: (x + y + z)n = [x + (y + z)]n] ⎛ 21 ⎞16 27. In the expansion of ⎜x −⎟ , the value of constant term is⎝ 2 ⎠x _________________ . 28. If the seventh terms from the beginning and the end in the expansion of ⎛ 31 ⎞ n 2 + are equal, then n equals _________________ .⎜ 3 ⎟⎝ 3⎠n −6 n −61( 1 )⎛ 1 ⎞6 n () 6 ⎛ 1 ⎞n 3[Hint : T = T ⇒ C2 = Cn−62 7n – 7 + 2 63 ⎜ 1 ⎟ ⎜ 1 ⎟ ⎜ ⎟ ⎜ 3 ⎟⎝ 33 ⎠ ⎝ 3 ⎠ n −12 n −12 ⎛⎞ 11 ⇒ (3 ) =⎜⎟1 ⇒ only problem when n – 12 = 0 ⇒ n = 12].2 33⎝⎠ ⎛12b⎞10 29. The coefficient of a– 6 b4 in the expansion of is _________.⎜− ⎟⎝a 3 ⎠1 b −2b 4 10 ⎛⎞⎛ ⎞ 1120 −64[Hint : T5 = C4 ⎜⎟⎜ ⎟= ab ]⎝⎠⎝ 3 27 a ⎠ 30. Middle term in the expansion of (a3 + ba)28 is _________ . 31. The ratio of the coefficients of xp andxq in the expansion of (1+x)p + q is_________ p + qC[Hint: p + q Cp = q] ⎛ x 3 ⎞10 32. The position of the term independent of x in the expansion of ⎜ +⎠ is2 ⎟⎝32x _________ . 33. 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 19 C1034. The sum of the series ∑20 Cr is 2 + r =02 35. The expression 79 + 97 is divisible by 64. Hint: 79 + 97 = (1 + 8)7 – (1 – 8)9 36. The number of terms in the expansion of [(2x + y3)4]7 is 8 37. The sum of coefficients of the two middle terms in the expansion of (1 + x)2n – 1 is equal to 2n – 1C n . 38. The last two digits of the numbers 3400 are 01. 12 n 39. If the expansion of ⎜⎛ x −2 ⎠⎟⎞ contains a term independent of x, then n is a⎝ x multiple of 2. 40. Number of terms in the expansion of (a + b)n where n ∈ Nis one less than the power n.

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