Check sibling questions

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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Transcript

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 is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.