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Example 15
Find the term independent of x in the expansion of (∛𝑥 " + " 1/(2 ∛𝑥))^18, x > 0.
Calculating general term of expansion
We know that general term of (a + b)n is
Tr+1 = nCr (a)n–r . (a)n
For general term of expansion (∛𝑥 " + " 1/(2 ∛𝑥))^18
Putting n = 18 , a = ∛𝑥 , b = 1/(2 ∛𝑥)
∴ Tr + 1 = 18Cr (∛𝑥)18 – r (1/(2 ∛𝑥))^𝑟
= 18Cr (〖𝑥")" 〗^(1/3 × (18 − 𝑟)) (1/2 "×" 1/∛𝑥)^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 ) (1/2)^𝑟 (1/∛𝑥)^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 ) 1/2^𝑟 (1/𝑥^(1/3) )^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 ) 1/2^𝑟 1/𝑥^(𝑟/3)
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 ) 1/2^𝑟 𝑥^((−𝑟)/3)
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 ) 𝑥^((−𝑟)/3) 1/2^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 𝑟)/3 − 𝑟/3) 1/2^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 𝑟 − 𝑟)/3 ) 1/2^𝑟
= 18Cr (〖𝑥")" 〗^((18 − 2𝑟)/3 ) 1/2^𝑟
∴ Tr + 1 = 18Cr (〖𝑥")" 〗^((18 − 2𝑟)/3 ) 1/2^𝑟
We need to find
the term independent of x
So, power of x is 0
𝑥^((18 −2𝑟)/3) = x0
Comparing power
(18 − 2𝑟)/3 = 0
18 – 2r = 0
18 = 2r
18/2 = r
9 = r
r = 9
Putting r = 9 in (1)
Tr+1 = 18Cr (〖𝑥")" 〗^((18 − 2𝑟)/3 ) 1/2^𝑟
T9+1 = 18C9 .𝑥^((18 −3(9))/3). 1/29
= 18C9 .x0. 1/29
= 18C9 . 1/29
Hence, the term which is independent of x is
10th term = T10 = 18C9 . 𝟏/𝟐𝟗a

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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