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Exercise - 8.2

Binomial Theorem

**Question-1 :-** Find the coefficient of x^{5} in (x + 3)^{8} .

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^{5}occurs in the (r + 1)th term of the expansion (x + 3)^{8}, we obtain T_{r+1}=^{8}C_{r}x^{8-r}3^{r}Comparing the indices of x in x^{5}and in T_{r+1}, we obtain r = 3 Thus, the coefficient of x^{5}is^{8}C_{3}3^{3}= 8!/3!5! x 3^{3}= 1512

**Question-2 :-** Find the coefficient of a^{5}b^{7} in (a – 2b)^{12}.

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 a^{5}b^{7}occurs in the (r + 1)th term of the expansion (a – 2b)^{12}, we obtain T_{r+1}=^{12}C_{r}a^{12-r}(-2b)^{r}=^{12}C_{r}(-2)^{r}a^{12-r}(b)^{r}Comparing the indices of a and b in a^{5}b^{7}and in T_{r+1}, we obtain r = 7 Thus, the coefficient of a^{5}b^{7}is^{12}C_{7}(-2)^{7}= 12!/7!5! x (-2)^{7}= (-792)(128) = -101376

**Question-3 :-** Write the general term in the expansion of (x^{2} – y)^{6}.

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}. Thus, the general term in the expansion of (x^{2}– y)^{6}is T_{r+1}=^{6}C_{r}(x^{2})^{6-r}(-y)^{r}=^{6}C_{r}(-1)^{r}x^{12-2r}y^{r}

**Question-4 :-** Write the general term in the expansion of (x^{2} – yx)^{12}, x ≠ 0.

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}. Thus, the general term in the expansion of (x^{2}– yx)^{12}is T_{r+1}=^{12}C_{r}(x^{2})^{12-r}(-yx)^{r}=^{12}C_{r}(-1)^{r}x^{24-2r}x^{r}y^{r}=^{12}C_{r}(-1)^{r}x^{24-r}y^{r}

**Question-5 :-** Find the 4th term in the expansion of (x – 2y)^{12}.

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}. Thus, the 4th term in the expansion of (x – 2y)^{12}is T_{4}= T_{3+1}=^{12}C_{3}x^{12-3}(-2y)^{3}= (-1)^{3}12!/3!9! x^{9}8y^{3}= -1760 x^{9}y^{3}

**Question-6 :-** Find the 13th term in the expansion of (9x - 1/3√x)^{18}, x ≠ 0.

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}. Thus, the 13th term in the expansion of (9x – 1/3√x)^{18}is T_{13}= T_{12+1}=^{18}C_{12}(9x)^{18-12}(-1/3√x)^{12}= (-1)^{12}18!/12!6! (9)^{6}(x)^{6}(1/3)^{12}(1/√x)^{12}= 18!/12!6! (3)^{12}(x)^{6}(1/3)^{12}(1/x)^{6}= 18!/12!6! = 18564

**Question-7 :-** Find the middle terms in the expansions of (3 - x^{3}/6)^{7}.

It is known that in the expansion of (a + b)^{n}, if n is odd, then there are two middle terms, namely, [(n+1)/2]th term and [(n+1)/2 + 1] term. Therefore, the middle terms in the expansion of (3 - x^{3}/6)^{7}are [(7+1)/2]th term and [(7+1)/2 + 1]th term. Now, T_{4}= T_{3+1}=^{7}C_{3}3^{7-3}(-x^{3}/6)^{3}= (-1)^{3}7!/3!4! 3^{4}(x^{9}/6^{3}) = -105x^{9}/8 Again, T_{5}= T_{4+1}=^{7}C_{4}3^{7-4}(-x^{3}/6)^{4}= (-1)^{4}7!/3!4! 3^{3}(x^{12}/6^{4}) = 35x^{12}/48 Thus, the middle terms in the expansion of (3 - x^{3}/6)^{7}are -105x^{9}/8 and 35x^{12}/48.

**Question-8 :-** Find the middle terms in the expansions of (x/3 + 9y)^{10}.

It is known that in the expansion of (a + b)^{n}, if n is even, then there are two middle terms, namely, [n/2 + 1]th term. Therefore, the middle terms in the expansion of (x/3 + 9y)^{10}are [10/2 + 1]th = 6th term. Now, T_{6}= T_{5+1}=^{10}C_{5}(x/3)^{10-5}(9y)^{5}= 9^{5}10!/5!5! x^{5}(1/3)^{5}y^{5}= 3^{10}10!/5!5! x^{5}1/3^{5}y^{5}= 61236 x^{5}y^{5}Thus, the middle terms in the expansion of (x/3 + 9y)^{10}is 61236 x^{5}y^{5}.

**Question-9 :-** In the expansion of (1 + a)^{m+n}, prove that coefficients of a^{m} and a^{n} 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 am occurs in the (r + 1)th term of the expansion (1 + a)^{m+n}, we obtain T^{r+1}=^{m+n}C_{r}1^{m+n-r}a^{r}=^{m+n}C_{r}a^{r}Comparing the indices of a in a^{m}and in T_{r+1}, we obtain r = m Therefore, the coefficient of a^{m}is^{m+n}C_{m}= (m+n)!/[m!(m+n-m)!] = (m+n)!/m!n! .....(i) Assuming that an occurs in the (k + 1)th term of the expansion (1 + a)^{m+n}, we obtain T^{k+1}=^{m+n}C_{k}1^{m+n-k}a^{k}=^{m+n}C_{k}a^{k}Comparing the indices of a in a^{n}and in T_{k+1}, we obtain k = n Therefore, the coefficient of a^{n}is^{m+n}C_{n}= (m+n)!/[n!(m+n-n)!] = (m+n)!/m!n! .....(ii) Thus, from (i) and (ii), it can be observed that the coefficients of a^{m}and a^{n}in the expansion of (1 + a)^{m+n}are equal.

**Question-10 :-** The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of (x + 1)^{n} are in the ratio 1 : 3 : 5. Find n and r.

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}. Therefore, (r – 1)th term in the expansion of (x + 1)^{n}is T_{r-1}=^{n}C_{r-2}x^{n-(r-2)}1^{r-2}=^{n}C_{r-2}x^{n-r+2}Now, rth term in the expansion of (x + 1)^{n}is T_{r}=^{n}C_{r-1}x^{n-(r-1)}1^{r-1}=^{n}C_{r-1}x^{n-r+1}Again, (r + 1)th term in the expansion of (x + 1)^{n}is T_{r+1}=^{n}C_{r}x^{n-r}1^{r}Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)^{n}are^{n}C_{r-2},^{n}C_{r-1}and^{n}C_{r}respectively. Since these coefficients are in the ratio 1:3:5, we obtain^{n}C_{r-2}:^{n}C_{r-1}= 1 : 3 n!/[(r-2)!(n-r+2)!] x [(r-1)!(n-r+1)!]/n! = 1/3 (r-1)/(n-r+2) = 1/3 n - r + 2 = 3r - 3 n - 4r + 5 = 0 .....(i)^{n}C_{r-1}:^{n}C_{r}= 3 : 5 n!/[(r-1)!(n-r+1)!] x [r!(n-r)!]/n! = 3/5 r/(n-r+1) = 3/5 3n - 3r + 3 = 5r 3n - 8r + 3 = 0 .....(ii) Multiplying (i) by 3 and subtracting it from (ii), we obtain 4r – 12 = 0, r = 3 Putting the value of r in (i), we obtain n – 12 + 5 = 0, n = 7 Thus, n = 7 and r = 3

**Question-11 :-** Prove that the coefficient of xn in the expansion of (1 + x)^{2n} is twice the coefficient of x^{n} in the expansion of (1 + x)^{2n-1}.

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^{n}occurs in the (r + 1)th term of the expansion of (1 + x)^{2n}, we obtain T_{r+1}=^{2n}C_{r}(1)^{2n-r}x^{r}=^{2n}C_{r}x^{r}Comparing the indices of x in x^{n}and in T_{r+1}, we obtain r = n Therefore, the coefficient of x^{n}in the expansion of (1 + x)^{2n}is^{2n}C_{n}= (2n)!/n!n! = (2n)!/(n!)^{2}.......(i) Assuming that x^{n}occurs in the (k + 1)th term of the expansion (1 + x)^{2n-1}, we obtain T_{k+1}=^{2n-1}C_{k}(1)^{2n-1-k}x^{k}=^{2n-1}C_{k}x^{k}Comparing the indices of x in x^{n}and T_{k+1}, we obtain k = n Therefore, the coefficient of x^{n}in the expansion of (1 + x)^{2n-1}is^{2n-1}C_{k}= (2n-1)!/[n!(n-1)!] = (2n)!/(2.n!n!) = 1/2 x [(2n)!/(n!)^{2}] .....(ii) From (i) and (ii), it is observed that 1/2 x^{2n}C_{n}=^{2n-1}C_{n}^{2n}C_{n}= 2.^{2n-1}C_{n}Therefore, the coefficient of x^{n}in the expansion of (1 + x)^{2n}is twice the coefficient of x^{n}in the expansion of (1 + x)^{2n-1}. Hence, proved.

**Question-12 :-** Find a positive value of m for which the coefficient of x^{2} in the expansion (1 + x)^{m} is 6.

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 of the expansion (1 + x)^{m}, we obtain T_{r+1}=^{m}C_{r}(1)^{m-r}x^{r}=^{m}C_{r}x^{r}Comparing the indices of x in x^{2}and in T_{r+1}, we obtain r = 2 Therefore, the coefficient of x^{2}is^{m}C_{2}. It is given that the coefficient of x2 in the expansion (1 + x)^{m}is 6. Therefore,^{m}C_{2}= 6 m!/[2!(m-2)!] = 6 [m x (m-1) x (m-2)!]/[2 x 1 x (m-2)!] = 6 [m(m-1)]/2 = 6 m^{2}- m = 12 m^{2}- m - 12 = 0 m^{2}- 4m + 3m - 12 = 0 m(m-4) + 3(m-4) = 0 (m + 3)(m - 4) = 0 m = -3 or m = 4 Thus, the positive value of m, for which the coefficient of x^{2}in the expansion (1 + x)^{m}is 6, is 4.

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