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Last updated at Jan. 29, 2020 by Teachoo
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Example 15 Find the term independent of x in the expansion of (โ๐ฅ " + " 1/(2 โ๐ฅ))^18, x > 0. Calculating general term of expansion We know that general term of (a + b)n is Tr+1 = nCr (a)nโr . (a)n For general term of expansion (โ๐ฅ " + " 1/(2 โ๐ฅ))^18 Putting n = 18 , a = โ๐ฅ , b = 1/(2 โ๐ฅ) โด Tr + 1 = 18Cr (โ๐ฅ)18 โ r (1/(2 โ๐ฅ))^๐ = 18Cr (ใ๐ฅ")" ใ^(1/3 ร (18 โ ๐)) (1/2 "ร" 1/โ๐ฅ)^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 ) (1/2)^๐ (1/โ๐ฅ)^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 ) 1/2^๐ (1/๐ฅ^(1/3) )^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 ) 1/2^๐ 1/๐ฅ^(๐/3) = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 ) 1/2^๐ ๐ฅ^((โ๐)/3) = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 ) ๐ฅ^((โ๐)/3) 1/2^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐)/3 โ ๐/3) 1/2^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ ๐ โ ๐)/3 ) 1/2^๐ = 18Cr (ใ๐ฅ")" ใ^((18 โ 2๐)/3 ) 1/2^๐ โด Tr + 1 = 18Cr (ใ๐ฅ")" ใ^((18 โ 2๐)/3 ) 1/2^๐ We need to find the term independent of x So, power of x is 0 ๐ฅ^((18 โ2๐)/3) = x0 Comparing power (18 โ 2๐)/3 = 0 18 โ 2r = 0 18 = 2r 18/2 = r 9 = r r = 9 Putting r = 9 in (1) Tr+1 = 18Cr (ใ๐ฅ")" ใ^((18 โ 2๐)/3 ) 1/2^๐ T9+1 = 18C9 .๐ฅ^((18 โ3(9))/3). 1/29 = 18C9 .x0. 1/29 = 18C9 . 1/29 Hence, the term which is independent of x is 10th term = T10 = 18C9 . ๐/๐๐a
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