TOPICS

Miscellaneous

Binomial Theorem

**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.

The (r + 1)th term of the expansion (a + b)^{n}is given by T_{r+1}=^{n}C_{r}a^{n-r}b^{r}. The first three terms of the expansion are given as 729, 7290, and 30375 respectively. Therefore, we obtain T_{1}=^{n}C_{0}a^{n-0}b^{0}=^{n}C_{0}a^{n}= a^{n}= 729 ....(i) T_{2}=^{n}C_{1}a^{n-1}b^{1}= na^{n-1}b = 7290 .....(ii) T_{3}=^{n}C_{2}a^{n-2}b^{2}= n(n-1)/2 . a^{n-2}. b^{2}= 30375 .....(iii) Dividing (ii) by (i), we obtain (na^{n-1}b)/a^{n}= 7290/729 nb/a = 10 .....(iv) Dividing (iii) by (ii), we obtain [n(n-1)/2 . a^{n-2}. b^{2}]/(na^{n-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 a^{6}= 729; a^{6}= 3^{6}; 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 x^{2} and x^{3} in the expansion of (3 + ax)^{9} are equal.

The (r + 1)th term of the expansion (x + y)^{n}is given by T_{r+1}=^{n}C_{r}x^{n-r}y^{r}. Assuming that x^{2}occurs in the (r + 1)th term in the expansion of (3 + ax)^{9}, we obtain T_{r+1}=^{9}C_{r}3^{9-r}(ax)^{r}=^{9}C_{r}3^{9-r}a^{r}x^{r}Comparing the indices of x in x^{2}and in T_{r+1}, we obtain r = 2 Thus, the coefficient of x^{2}is T_{2+1}=^{9}C_{2}3^{9-2}a^{2}= 36 3^{7}a^{2}Assuming that x^{3}occurs in the (k + 1)th term in the expansion of (3 + ax)^{9}, we obtain T_{k+1}=^{9}C_{k}3^{9-k}(ax)^{k}=^{9}C_{k}3^{9-k}a^{k}x^{k}Comparing the indices of x in x^{3}and in T_{k+1}, we obtain k = 3 Thus, the coefficient of x^{3}is T_{3+1}=^{9}C_{3}3^{9-3}a^{3}= 84 3^{6}a^{3}It is given that the coefficients of x^{2}and x^{3}are the same. 84 3^{6}a^{3}= 36 3^{7}a^{2}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 x^{5} in the product (1 + 2x)^{6} (1 – x)^{7} using binomial theorem.

(1 + 2x)^{6}=^{6}C_{0}(1)^{6}+^{6}C_{1}(1)^{5}(2x) +^{6}C_{2}(1)^{4}(2x)^{2}+^{6}C_{3}(1)^{3}(2x)^{3}+^{6}C_{4}(1)^{2}(2x)^{4}+^{6}C_{5}(1)(2x)^{5}+^{6}C_{6}(2x)^{6}= 1 + 6(2x) + 15(2x)^{2}+ 20(2x^{3}+ 15(2x)^{4}+ 6(2x)^{5}+ (2x)^{6}= 1 + 12x + 60x^{2}+ 160x^{3}+ 240x^{4}+ 192x^{5}+ 64x^{6}(1 - x)^{7}=^{7}C_{0}(1)^{7}-^{7}C_{1}(1)^{6}(x) +^{7}C_{2}(1)^{5}(x)^{2}-^{7}C_{3}(1)^{4}(x)^{3}+^{7}C_{4}(1)^{3}(x)^{4}-^{7}C_{5}(1)^{2}(x)^{5}+^{7}C_{6}(1)(x)^{6}-^{7}C_{7}(x)^{7}= 1 - 7x + 21x^{2}- 35x^{3}+ 35x^{4}- 21x^{5}+ 7x^{6}- x^{7}Therefore, (1 + 2x)^{6}. (1 - x)^{7}= (1 + 12x + 60x^{2}+ 160x^{3}+ 240x^{4}+ 192x^{5}+ 64x^{6}) . (1 - 7x + 21x^{2}- 35x^{3}+ 35x^{4}- 21x^{5}+ 7x^{6}- x^{7}) The complete multiplication of the two brackets is not required to be carried out. Only those terms, which involve x^{5}, are required. The terms containing x^{5}are 1(-21x^{5}) + (12x)(35x^{4}) + (60x^{2})(-35x^{3}) + (160x^{3})(21x^{2}) + (240x^{4})(-7x) + (192x^{5})(1) = 171x^{5}Thus, the coefficient of x^{5}in the given product is 171.

**Question-4 :-** If a and b are distinct integers, prove that a – b is a factor of a^{n} – b^{n}, whenever n is a positive integer.

In order to prove that (a – b) is a factor of (a^{n}– b^{n}), it has to be proved that a^{n}– b^{n}= 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 (a^{n}– b^{n}), where n is a positive integer.

**Question-5 :-** Evaluate (√3 + √2)^{6} - (√3 - √2)^{6} .

(a + b)^{6}=^{6}C_{0}(a)^{6}+^{6}C_{1}(a)^{5}(b) +^{6}C_{2}(a)^{4}(b)^{2}+^{6}C_{3}(a)^{3}(b)^{3}+^{6}C_{4}(a)^{2}(b)^{4}+^{6}C_{5}(a)(b)^{5}+^{6}C_{6}(b)^{6}(a - b)^{6}=^{6}C₀(a)^{6}-^{6}C_{1}(a)^{5}(b) +^{6}C_{2}(a)^{4}(b)^{2}-^{6}C₃(a)^{3}(b)^{3}+^{6}C_{4}(a)^{2}(b)^{4}-^{6}C_{5}(a)(b)^{5}+^{6}C_{6}(b)^{6}Now, (a + b)^{6}- (a - b)^{6}=^{6}C_{0}(a)^{6}+^{6}C_{1}(a)^{5}(b) +^{6}C_{2}(a)^{4}(b)^{2}+^{6}C_{3}(a)^{3}(b)^{3}+^{6}C_{4}(a)^{2}(b)^{4}+^{6}C_{5}(a)(b)^{5}+^{6}C_{6}(b)^{6}- [^{6}C_{0}(a)^{6}-^{6}C_{1}(a)^{5}(b) +^{6}C_{2}(a)^{4}(b)^{2}-^{6}C_{3}(a)^{3}(b)^{3}+^{6}C_{4}(a)^{2}(b)^{4}-^{6}C_{5}(a)(b)^{5}+^{6}C_{6}(b)^{6}] =^{6}C_{0}(a)^{6}+^{6}C_{1}(a)^{5}(b) +^{6}C_{2}(a)^{4}(b)^{2}+^{6}C_{3}(a)^{3}(b)^{3}+^{6}C_{4}(a)^{2}(b)^{4}+^{6}C_{5}(a)(b)^{5}+^{6}C_{6}(b)^{6}-^{6}C_{0}(a)^{6}+^{6}C_{1}(a)^{5}(b) -^{6}C_{2}(a)^{4}(b)^{2}+^{6}C_{3}(a)^{3}(b)^{3}-^{6}C_{4}(a)^{2}(b)^{4}+^{6}C_{5}(a)(b)^{5}-^{6}C_{6}(b)^{6}= 2(^{6}C_{1}(a)^{5}(b) +^{6}C_{3}(a)^{3}(b)^{3}+^{6}C_{5}(a)(b)^{5}) = 2(6a^{5}b + 20a^{3}b^{3}+ 6ab^{5}) 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 :-** .

(x + y)^{4}=^{4}C_{0}(x)^{4}+^{4}C_{1}(x)^{3}.y +^{4}C_{2}(x)^{2}.y^{2}+^{4}C_{3}(x).y^{3}+^{4}C_{4}.y^{4}(x - y)^{4}=^{4}C_{0}(x)^{4}-^{4}C_{1}(x)^{3}.y +^{4}C_{2}(x)^{2}.y^{2}-^{4}C_{3}(x).y^{3}+^{4}C_{4}.y^{4}Now, (x + y)^{4}+ (x - y)^{4}=^{4}C_{0}(x)^{4}+^{4}C_{1}(x)^{3}.y +^{4}C_{2}(x)^{2}.y^{2}+^{4}C_{3}(x).y^{3}+^{4}C_{4}.y^{4}+^{4}C_{0}(x)^{4}-^{4}C_{1}(x)^{3}.y +^{4}C_{2}(x)^{2}.y^{2}-^{4}C_{3}(x).y^{3}+^{4}C_{4}.y^{4}= 2(^{4}C_{0}(x)^{4}+^{4}C_{2}(x)^{2}.(y)^{2}+^{4}C_{4}.(y)^{4}) = 2(x^{4}+ 6x^{2}y^{2}+ y^{4}) By putting x = a^{2}and y = √a^{2}- 1, we obtain

**Question-7 :-** Find an approximation of (0.99)^{5} using the first three terms of its expansion.

0.99 = 1 – 0.01 (0.99)^{5}= (1 - 0.01)^{5}=^{5}C_{0}(1)^{5}-^{5}C_{1}(1)^{4}(0.01) +^{5}C_{2}(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

**Question-9 :-** Expand using Binomial Theorem (1 + x/2 - 2/x)^{4}, x ≠ 0.

**Question-10 :-** Find the expansion of (3x^{2} – 2ax + 3a^{2})^{3} using binomial theorem.

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