Find the middle terms in the expansion of-i 3 x-x 369ii 2 x 2-1 x 7iii 3 x-2 - 215iv x4-1x311

(i) Here, n, i.e. 9, is an odd number. Thus, the middle terms are n+12th #160;and #160;n+12+1th, #160;i.e. #160;5th #160;and #160;6th Now,T5=T4+1=C4 ...

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Byju's Answer

Standard XI

Mathematics

Sequence

Find the midd...

Question

Find the middle terms in the expansion of:
(i) ${(3 x - \frac{x^{3}}{6})}^{9}$

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

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

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

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Solution

(i) Here, n, i.e. 9, is an odd number.
Thus, the middle terms are $(\frac{n + 1}{2}) th and (\frac{n + 1}{2} + 1) th, i . e . 5 th and 6 th$
$Now, T_{5} = T_{4 + 1} = {}^{9}C_{4} (3 x)^{9 - 4} {(\frac{- x^{3}}{6})}^{4} = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 27 \times 9 \times \frac{1}{36 \times 36} x^{17} = \frac{189}{8} x^{17} and, T_{6} = T_{5 + 1} = {}^{9}C_{5} (3 x)^{9 - 5} {(\frac{- x^{3}}{6})}^{5} = - \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2} \times 81 \times \frac{1}{216 \times 36} x^{19} = - \frac{21}{16} x^{19}$

(ii) Here, n, i.e., 7, is an odd number.

$Thus, the middle terms are (\frac{7 + 1}{2}) th and (\frac{7 + 1}{2} + 1) th i . e . 4 th and 5 th Now, T_{4} = T_{3 + 1} = {}^{7}C_{3} (2 x^{2})^{7 - 3} {(\frac{- 1}{x})}^{3} = - \frac{7 \times 6 \times 5}{3 \times 2} \times 16 x^{8} \times \frac{1}{x^{3}} = - 560 x^{5} And, T_{5} = T_{4 + 1} = {}^{7}C_{4} (2 x^{2})^{7 - 4} {(\frac{- 1}{x})}^{4} = 35 \times 8 \times x^{6} \times \frac{1}{x^{4}} = 280 x^{2}$

(iii)
$Given : n, i . e . 15 is an odd number . Thus, the middle terms are (\frac{15 + 1}{2}) th and (\frac{15 + 1}{2} + 1) th i . e . 8 th and 9 th . Now, T_{8} = T_{7 + 1} = {}^{15}C_{7} (3 x)^{15 - 7} {(\frac{- 2}{x^{2}})}^{7} = - \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^{8} \times 2^{7} x^{8 - 14} = \frac{- 6435 \times 3^{8} \times 2^{7}}{x^{6}} And, T_{9} = T_{8 + 1} = {}^{15}C_{8} (3 x)^{15 - 8} {(\frac{- 2}{x^{2}})}^{8} = \frac{15 \times 14 \times 13 \times 12 \times 11 \times 10 \times 9}{7 \times 6 \times 5 \times 4 \times 3 \times 2} \times 3^{7} \times 2^{8} \times x^{7 - 16} = \frac{6435 \times 3^{7} \times 2^{8}}{x^{9}}$

(iv)
$Here, n, i . e ., 11, is an odd number . Thus, the middle terms are (\frac{11 + 1}{2}) th and (\frac{11 + 1}{2} + 1) th i . e . 6 th and 7 th . Now, T_{6} = T_{5 + 1} = {}^{11}C_{5} (x^{4})^{11 - 5} {(\frac{- 1}{x^{3}})}^{5} = - \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} \times {(x)}^{24 - 15} = - 462 x^{9} And, T_{7} = T_{6 + 1} = {}^{11}C_{6} (x^{4})^{11 - 6} {(\frac{- 1}{x^{3}})}^{6} = \frac{11 \times 10 \times 9 \times 8 \times 7}{5 \times 4 \times 3 \times 2} {(x)}^{20 - 18} = 462 x^{2}$

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Similar questions

Find the middle term in the expansion of:

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

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

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

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Find the middle term in the expansion of:

$(i) {(\frac{2}{3} x - \frac{3}{2 x})}^{20}$

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Middle term in the expansion of $(1 + 3 x + 3 x^{2} + x^{3})^{6}$ is

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Q. Find the middle terms(s) in the expansion of:
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Find the middle terms in the expansion of: (i) 3x-x369 (ii) 2x2-1x7 (iii) 3x-2x215 (iv) x4-1x311

Find the middle terms in the expansion of:
(i) ${(3 x - \frac{x^{3}}{6})}^{9}$

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

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

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