Question-1 :- Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.
Solution :-The (r + 1)th term of the expansion (a + b)n is given by Tr+1 = nCr an-r br. The first three terms of the expansion are given as 729, 7290, and 30375 respectively. Therefore, we obtain T1 = nC0 an-0 b0 = nC0 an = an = 729 ....(i) T2 = nC1 an-1 b1 = nan-1 b = 7290 .....(ii) T3 = nC2 an-2 b2 = n(n-1)/2 . an-2 . b2 = 30375 .....(iii) Dividing (ii) by (i), we obtain (nan-1 b)/an = 7290/729 nb/a = 10 .....(iv) Dividing (iii) by (ii), we obtain [n(n-1)/2 . an-2 . b2]/(nan-1 b) = 30375/7290 [(n-1)b]/2a = 30375/7290 [(n-1)b]/a = 25/3 nb/a - b/a = 25/3 10 - b/a = 25/3 [using (iv)] b/a = 10 - 25/3 b/a = 5/3 .....(v) From (iv) and (v), we obtain n x 5/3 = 10 n = 30/5 n = 6 Substituting n = 6 in equation (i), we obtain a6 = 729; a6 = 36; a = 3 From (v), we obtain b/3 = 5/3 b = 5 Thus, a = 3, b = 5, and n = 6.
Question-2 :- Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.
Solution :-The (r + 1)th term of the expansion (x + y)n is given by Tr+1 = nCr xn-r yr. Assuming that x2 occurs in the (r + 1)th term in the expansion of (3 + ax)9, we obtain Tr+1 = 9Cr 39-r (ax)r = 9Cr 39-r ar xr Comparing the indices of x in x2 and in Tr+1, we obtain r = 2 Thus, the coefficient of x2 is T2+1 = 9C2 39-2 a2 = 36 37 a2 Assuming that x3 occurs in the (k + 1)th term in the expansion of (3 + ax)9, we obtain Tk+1 = 9Ck 39-k (ax)k = 9Ck 39-k ak xk Comparing the indices of x in x3 and in Tk+1, we obtain k = 3 Thus, the coefficient of x3 is T3+1 = 9C3 39-3 a3 = 84 36 a3 It is given that the coefficients of x2 and x3 are the same. 84 36 a3 = 36 37 a2 84a = 36 x 3 a = 108/84 a = 9/7 Thus, the required value of a is 9/7.
Question-3 :- Find the coefficient of x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem.
Solution :-(1 + 2x)6 = 6 C0(1)6 + 6 C1(1)5 (2x) + 6 C2(1)4 (2x)2 + 6 C3(1)3 (2x)3 + 6 C4(1)2 (2x)4 + 6 C5(1)(2x)5 + 6 C6(2x)6 = 1 + 6(2x) + 15(2x)2 + 20(2x3 + 15(2x)4 + 6(2x)5 + (2x)6 = 1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6 (1 - x)7 = 7C0(1)7 - 7C1(1)6(x) + 7C2(1)5(x)2 - 7C3(1)4(x)3 + 7C4(1)3(x)4 - 7C5(1)2(x)5 + 7C6(1)(x)6 - 7C7(x)7 = 1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7 Therefore, (1 + 2x)6 . (1 - x)7 = (1 + 12x + 60x2 + 160x3 + 240x4 + 192x5 + 64x6) . (1 - 7x + 21x2 - 35x3 + 35x4 - 21x5 + 7x6 - x7) The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve x5, are required. The terms containing x5 are 1(-21x5) + (12x)(35x4) + (60x2)(-35x3) + (160x3)(21x2) + (240x4)(-7x) + (192x5)(1) = 171x5 Thus, the coefficient of x5 in the given product is 171.
Question-4 :- If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer.
Solution :-In order to prove that (a – b) is a factor of (an – bn), it has to be proved that an – bn = k (a – b), where k is some natural number It can be written that, a = a – b + b This shows that (a – b) is a factor of (an – bn), where n is a positive integer.
Question-5 :- Evaluate (√3 + √2)6 - (√3 - √2)6 .
Solution :-(a + b)6 = 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6 (a - b)6 = 6C₀(a)6 - 6C1(a)5(b) + 6C2(a)4(b)2 - 6C₃(a)3(b)3 + 6C4(a)2(b)4 - 6C5(a)(b)5 + 6C6(b)6 Now, (a + b)6 - (a - b)6 = 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6 - [6C0(a)6 - 6C1(a)5(b) + 6C2(a)4(b)2 - 6C3(a)3(b)3 + 6C4(a)2(b)4 - 6C5(a)(b)5 + 6C6(b)6] = 6C0(a)6 + 6C1(a)5(b) + 6C2(a)4(b)2 + 6C3(a)3(b)3 + 6C4(a)2(b)4 + 6C5(a)(b)5 + 6C6(b)6 - 6C0(a)6 + 6C1(a)5(b) - 6C2(a)4(b)2 + 6C3(a)3(b)3 - 6C4(a)2(b)4 + 6C5(a)(b)5 - 6C6(b)6 = 2(6C1(a)5(b) + 6C3(a)3(b)3 + 6C5(a)(b)5) = 2(6a5b + 20a3b3 + 6ab5) By putting a = √3 and b = √2, we obtain = 2(6(√3)5(√2) + 20(√3)3(√2)3 + 6(√3)(√2)5) = 2(54√6 + 120√6 + 24√6) = 2 x 198 x √6 = 396√6
Question-6 :- 
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(x + y)4 = 4C0(x)4 + 4C1(x)3.y + 4C2(x)2.y2 + 4C3(x).y3 + 4C4.y4 (x - y)4 = 4C0(x)4 - 4C1(x)3.y + 4C2(x)2.y2 - 4C3(x).y3 + 4C4.y4 Now, (x + y)4 + (x - y)4 = 4C0(x)4 + 4C1(x)3.y + 4C2(x)2.y2 + 4C3(x).y3 + 4C4.y4 + 4C0(x)4 - 4C1(x)3.y + 4C2(x)2.y2 - 4C3(x).y3 + 4C4.y4 = 2(4C0(x)4 + 4C2(x)2.(y)2 + 4C4.(y)4) = 2(x4 + 6x2y2 + y4) By putting x = a2 and y = √a2 - 1, we obtain
Question-7 :- Find an approximation of (0.99)5 using the first three terms of its expansion.
Solution :-0.99 = 1 – 0.01 (0.99)5 = (1 - 0.01)5 = 5C0 (1)5 - 5C1 (1)4 (0.01) + 5C2 (1)3 (0.01)2 [approx.] = 1 - 5(0.01) + 10(0.01)2 = 1 - 0.05 + 0.001 = 0.001 - 0.05 = 0.951 Thus, the value of (0.99)5 is approximately 0.951.
Question-8 :- Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of
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Question-9 :- Expand using Binomial Theorem (1 + x/2 - 2/x)4, x ≠ 0.
Solution :-
Question-10 :- Find the expansion of (3x2 – 2ax + 3a2)3 using binomial theorem.
Solution :-
