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(x±a)(x±b) = x²±(a+b)x+ab.

Find the midd...

Question

Find the middle terms(s) in the expansion of:
(i) ${(x - \frac{1}{x})}^{10}$

(ii) ${(1 - 2 x + x^{2})}^{n}$

(iii) ${(1 + 3 x + 3 x^{2} + x^{3})}^{2 n}$

(iv) ${(2 x - \frac{x^{2}}{4})}^{9}$

(v) ${(x - \frac{1}{x})}^{2 n + 1}$

(vi) ${(\frac{x}{3} + 9 y)}^{10}$

(vii) ${(3 - \frac{x^{3}}{6})}^{7}$

(viii) ${(2 a x - \frac{b}{x^{2}})}^{12}$

(ix) ${(\frac{p}{x} + \frac{x}{p})}^{9}$

(x) ${(\frac{x}{a} - \frac{a}{x})}^{10}$

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Solution

(i)
${(x - \frac{1}{x})}^{10} Here, n is an even number . ∴ Middle term = (\frac{10}{2} + 1) th = 6 th term Now, we have T_{6} = T_{5 + 1} = {}^{10}C_{5} x^{10 - 5} {(\frac{- 1}{x})}^{5} = - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} = - 252$

(ii)
$(1 - 2 x + x^{2})^{n} = (1 - x)^{2 n} n is an even number . ∴ Middle term = (\frac{2 n}{2} + 1) th = (n + 1) th term Now, we have T_{n + 1} = {}^{2 n}C_{n} (- 1)^{n} (x)^{n} = \frac{(2 n)!}{(n!)^{2}} (- 1)^{n} x^{n}$

(iii)
$(1 + 3 x + 3 x^{2} + x^{3})^{2 n} = (1 + x)^{6 n} Here, n is an even number . ∴ Middle term = (\frac{6 n}{2} + 1) th = (3 n + 1) th term Now, we have T_{3 n + 1} = {}^{6 n}C_{3 n} x^{3 n} = \frac{(6 n)!}{(3 n!)^{2}} x^{3 n}$

(iv)
${(2 x - \frac{x^{2}}{4})}^{9} Here, n is an odd number . Therefore, the middle terms are (\frac{n + 1}{2}) th and (\frac{n + 1}{2} + 1) th, i . e . 5 th and 6 th terms . Now, we have T_{5} = T_{4 + 1} = {}^{9}C_{4} (2 x)^{9 - 4} {(\frac{- x^{2}}{4})}^{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^{5} \frac{1}{4^{4}} x^{5 + 8} = \frac{63}{4} x^{13} And, T_{6} = T_{5 + 1} = {}^{9}C_{5} (2 x)^{9 - 5} {(\frac{- x^{2}}{4})}^{5} = - \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 2^{4} \frac{1}{4^{5}} x^{4 + 10} = - \frac{63}{32} x^{14}$

(v)
${(x - \frac{1}{x})}^{2 n + 1} Here, (2 n + 1) is an odd number . Therefore, the middle terms are (\frac{2 n + 1 + 1}{2}) th and (\frac{2 n + 1 + 1}{2} + 1) th i . e . (n + 1) th and (n + 2) th terms . Now, we have : T_{n + 1} = {}^{2 n + 1}C_{n} x^{2 n + 1 - n} \times \frac{(- 1)^{n}}{x^{n}} = (- 1)^{n} {}^{2 n + 1}C_{n} x And, T_{n + 2} = T_{n + 1 + 1} = {}^{2 n + 1}C_{n + 1} x^{2 n + 1 - n - 1} \frac{(- 1)^{n + 1}}{x^{n + 1}} = (- 1)^{n + 1} {}^{2 n + 1}C_{n + 1} \times \frac{1}{x}$

(vi)
${(\frac{x}{3} + 9 y)}^{10} Here, n is an even number . Therefore, the middle term is (\frac{10}{2} + 1) th, i . e ., 6 th term . Now, we have T_{6} = T_{5 + 1} = {}^{10}C_{5} {(\frac{x}{3})}^{10 - 5} (9 y)^{5} = \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2} \times \frac{1}{3^{5}} \times 9^{5} \times x^{5} y^{5} = 61236 x^{5} y^{5}$

(vii)
${(3 - \frac{x^{3}}{6})}^{7} Here, n is an odd number . Therefore, the middle terms are (\frac{7 + 1}{2}) th and (\frac{7 + 1}{2} + 1) th, i . e ., 4 th and 5 th terms . Now, we have T_{4} = T_{3 + 1} = {}^{7}C_{3} {(3)}^{7 - 3} {(\frac{- x^{3}}{6})}^{3} = - \frac{105}{8} x^{9} And, T_{5} = T_{4 + 1} = {}^{9}C_{4} {(3)}^{9 - 4} {(\frac{- x^{3}}{6})}^{4} = \frac{7 \times 6 \times 5}{3 \times 2} \times 3^{5} \times \frac{1}{6^{4}} x^{12} = \frac{35}{48} x^{12}$

(viii)
${(2 a x - \frac{b}{x^{2}})}^{12} Here, n is an even number . ∴ Middle term = {(\frac{12}{2} + 1)}^{th} = 7^{th} term Now, we have T_{7} = T_{6 + 1} = {}^{12}C_{6} {(2 a x)}^{12 - 6} {(\frac{- b}{x^{2}})}^{6} = \frac{12 \times 11 \times 10 \times 9 \times 8 \times 7}{6 \times 5 \times 4 \times 3 \times 2 \times 1} \times {(\frac{2 a b}{x})}^{6} = \frac{59136 a^{6} b^{6}}{x^{6}}$

(ix)
${(\frac{p}{x} + \frac{x}{p})}^{9} Here, n is an odd number . Therefore, the middle terms are {(\frac{9 + 1}{2})}^{th} and {(\frac{9 + 1}{2} + 1)}^{th}, i . e ., 5^{th} and 6^{th} terms . Now, we have T_{5} = T_{4 + 1} = {}^{9}C_{4} {(\frac{p}{x})}^{9 - 4} {(\frac{x}{p})}^{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times (\frac{p}{x}) = \frac{126 p}{x} And, T_{6} = T_{5 + 1} = {}^{9}C_{5} {(\frac{p}{x})}^{9 - 5} {(\frac{x}{p})}^{5} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} \times (\frac{x}{p}) = \frac{126 x}{p}$

(x)
${(\frac{x}{a} - \frac{a}{x})}^{10} Here, n is an even number . ∴ Middle term = {(\frac{10}{2} + 1)}^{th} = 6^{th} term Now, we have T_{6} = T_{5 + 1} = {}^{10}C_{5} {(\frac{x}{a})}^{10 - 5} {(\frac{- a}{x})}^{5} = - \frac{10 \times 9 \times 8 \times 7 \times 6}{5 \times 4 \times 3 \times 2 \times 1} = - 252$

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Similar questions

Find the middle term(s)in the expansion of:

$(i) {(x - \frac{1}{x})}^{10}$

$(i i) (1 - 2 x + x^{2})^{n}$

$(i i i) (1 + 3 x + 3 x^{2} + x^{3})^{2 n}$

$(i v) {(2 x - \frac{x^{2}}{4})}^{9}$

$(v) {(x - \frac{1}{x})}^{2 n + 1}$

$(v i) {(\frac{x}{3} + 9 y)}^{10}$

$(v i i) {(3 - \frac{x^{3}}{6})}^{7}$

$(v i i i) {(2 a x - \frac{b}{x^{2}})}^{12}$

$(i x) {(\frac{p}{x} + \frac{x}{p})}^{9}$

$(x) {(\frac{x}{a} - \frac{a}{x})}^{10}$

Find the middle term in the expansion of:

$(i) {(3 x - \frac{x^{3}}{6})}^{9}$

$(i i) {(2 x^{2} - \frac{1}{x})}^{7}$

$(i i i) {(3 x - \frac{2}{x^{2}})}^{15}$

$(i v) {(x^{4} - \frac{1}{x^{3}})}^{11}$

Now in each of the following, find the quotient and remainder on dividing p(x) by q(x) and write then in the form p(x) = u(x) q(x) + v(x).

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

Find the cube of $x + \frac{1}{x}$

Q. Question 14
By Remainder theorem, find the remainder when p(x) is divided by g(x).

(i)

p (x) = x^{3} - 2 x^{2} - 4 x - 1, g (x) = x + 1

p (x) = x^{3} - 3 x^{2} + 4 x + 50, g (x) = x - 3

p (x) = 4 x^{3} - 12 x^{2} + 14 x - 3, g (x) = 2 x - 1

p (x) = x^{3} - 6 x^{2} + 2 x - 4, g (x) = 1 - \frac{3}{2} x

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