Find middle term- 3-x3-67

( 3 x^3/7 )^7 It will have 8 terms ∴there will be two middle terms. T_4 and T_5 T_4 = ^7C_3(3)^7 3 ( x^3/6 )^3 = 105/8x^9 T_5=^7C_4(3)^7 4 ( x^3/6 )^4 = 35/48 ...

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Standard XII

Mathematics

General Term of Binomial Expansion

Find middle t...

Question

Find middle term: $(3 - x^{\frac{3}{6}})^{7}$

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Solution

${(3 - \frac{x^{3}}{7})}^{7}$
It will have 8 terms \therefore there will be two middle terms.
$T_{4}$ and $T_{5}$
$T_{4} =^{7} C_{3} (3)^{7 - 3} {(\frac{- x^{3}}{6})}^{3} = \frac{- 105}{8} x^{9}$
$T_{5} =^{7} C_{4} (3)^{7 - 4} {(\frac{- x^{3}}{6})}^{4} = \frac{35}{48} x^{12}$

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Find middle term: (3−x36)7

(3−x37)7 It will have 8 terms \therefore there will be two middle terms. T4 and T5 T4=7C3(3)7−3(−x36)3=−1058x9 T5=7C4(3)7−4(−x36)4=3548x12

Find middle term: $(3 - x^{\frac{3}{6}})^{7}$

${(3 - \frac{x^{3}}{7})}^{7}$
It will have 8 terms \therefore there will be two middle terms.
$T_{4}$ and $T_{5}$
$T_{4} =^{7} C_{3} (3)^{7 - 3} {(\frac{- x^{3}}{6})}^{3} = \frac{- 105}{8} x^{9}$
$T_{5} =^{7} C_{4} (3)^{7 - 4} {(\frac{- x^{3}}{6})}^{4} = \frac{35}{48} x^{12}$