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Sequence

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

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

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