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Example 15 - Find term independent of x in the expansion (x^1/3 + 1/2x

Example 15 - Chapter 8 Class 11 Binomial Theorem - Part 2
Example 15 - Chapter 8 Class 11 Binomial Theorem - Part 3 Example 15 - Chapter 8 Class 11 Binomial Theorem - Part 4 Example 15 - Chapter 8 Class 11 Binomial Theorem - Part 5

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Example 15 Find the term independent of x in the expansion of (βˆ›π‘₯ " + " 1/(2 βˆ›π‘₯))^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 (βˆ›π‘₯ " + " 1/(2 βˆ›π‘₯))^18 Putting n = 18 , a = βˆ›π‘₯ , b = 1/(2 βˆ›π‘₯) ∴ Tr + 1 = 18Cr (βˆ›π‘₯)18 – r (1/(2 βˆ›π‘₯))^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^(1/3 Γ— (18 βˆ’ π‘Ÿ)) (1/2 "Γ—" 1/βˆ›π‘₯)^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 ) (1/2)^π‘Ÿ (1/βˆ›π‘₯)^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 ) 1/2^π‘Ÿ (1/π‘₯^(1/3) )^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 ) 1/2^π‘Ÿ 1/π‘₯^(π‘Ÿ/3) = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 ) 1/2^π‘Ÿ π‘₯^((βˆ’π‘Ÿ)/3) = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 ) π‘₯^((βˆ’π‘Ÿ)/3) 1/2^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ)/3 βˆ’ π‘Ÿ/3) 1/2^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ π‘Ÿ βˆ’ π‘Ÿ)/3 ) 1/2^π‘Ÿ = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ 2π‘Ÿ)/3 ) 1/2^π‘Ÿ ∴ Tr + 1 = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ 2π‘Ÿ)/3 ) 1/2^π‘Ÿ We need to find the term independent of x So, power of x is 0 π‘₯^((18 βˆ’2π‘Ÿ)/3) = x0 Comparing power (18 βˆ’ 2π‘Ÿ)/3 = 0 18 – 2r = 0 18 = 2r 18/2 = r 9 = r r = 9 Putting r = 9 in (1) Tr+1 = 18Cr (γ€–π‘₯")" γ€—^((18 βˆ’ 2π‘Ÿ)/3 ) 1/2^π‘Ÿ T9+1 = 18C9 .π‘₯^((18 βˆ’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/πŸπŸ—a

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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.