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  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise

Transcript

Misc 8 Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of (∜2+ 1/∜3)^𝑛 is √6 : 1 We know that General term of expansion (a + b)n Tr + 1 = nCr an – rbr Fifth term from beginning We need to calculate T5 = T4 + 1 Putting r = 4 , a = ∜2 , b = (1/∜3) T4 + 1 = nC4(∜2)n – 4.(1/∜3)4 T5 = nC4 γ€–((2)γ€—^(1/4))n – 4 .(1/3^(1/4) )4 T5 = nC4 γ€–(2)γ€—^(1/4 Γ— (𝑛 βˆ’ 4)) .(1/3^(1/4) )4 = nCr γ€–(2)γ€—^(1/4 Γ—(π‘›βˆ’4)) .(1/3) Finding 5th term from end We know that (a + b)n = nCo an bo +nC1 an–1 b1 +……..+ nCn–1 (a)n – (n–1) .bn–1 +nCn a0 bn = an + nC1 an–1b1 + ………………………+ nC1 a1bn–1 + bn = bn + nC1 a1 bn–1 +…………………+ nC1 an–1 b1 + an Hence, 5th term from end = (n – 3)th term from beginning Another method, rth term from end = (n – r + 2)th term from stating 5th term from end = (n – 5 + 2)th term from stating = (n – 3)th term from beginning Now we need to calculate (n – 3)th term of expansion (√(πŸ’&𝟐) + 𝟏/√(πŸ’&πŸ‘) )r Tr + 1 = nCr an – rbr Putting r = (n – 3) – 1 = n – 4 a = ∜2 , b = 1/∜3 T(n – 4 + 1) = nCn – 4 (∜2)^(𝑛 βˆ’(π‘›βˆ’4)) . (1/∜3)^(𝑛 βˆ’ 4) = nCn – 4 [γ€–(2)γ€—^(1/4) ]^"n – n + 4" . (1/3^(1/4) )^(𝑛 βˆ’ 4) = nCn – 4 [γ€–(2)γ€—^(1/4) ]^"4" . (3^((βˆ’1)/4) )^(𝑛 βˆ’ 4) = nCn – 4 γ€–(2)γ€—^1 .(1/3)^((𝑛 βˆ’ 4)/4) = nCn – 4 γ€–(2)γ€—^(4/4) .3^((βˆ’(𝑛 βˆ’ 4))/4) = nCn – 4 γ€–(2)γ€—^1 .(1/3)^((𝑛 βˆ’ 4)/4) Given that Ratio fifth term from beginning fifth term from end is √6 : 1 (πΉπ‘–π‘“π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š π‘“π‘Ÿπ‘œπ‘š 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 )/(πΉπ‘–π‘“π‘‘β„Ž π‘‘π‘’π‘Ÿπ‘š π‘“π‘Ÿπ‘œπ‘š 𝑒𝑛𝑑 ) = √6/1 (〖𝑛𝐢〗_4 2^((π‘›βˆ’4)/4) . 1/3)/(〖𝑛𝐢〗_(π‘›βˆ’4) 2 . (1/3)^((π‘›βˆ’4)/4) ) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ) . (2)^((𝑛 βˆ’ 4)/4 βˆ’1 ) (1/3)^( 1 βˆ’ ((𝑛 βˆ’ 4)/4) ) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ) . 2^((𝑛 βˆ’ 4 βˆ’ 4)/4 ). (1/3)^((4 βˆ’ 𝑛 + 4)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((𝑛 βˆ’ 8)/4 ). (1/3)^((8 βˆ’ 𝑛)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((𝑛 βˆ’ 8)/4 ). (1/3)^((βˆ’(𝑛 βˆ’ 8))/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 2^((π‘›βˆ’8)/4). (3)^((𝑛 βˆ’ 8)/4) = √6/1 (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). γ€–(2 Γ— 3)γ€—^((𝑛 βˆ’ 8)/4). = 6^(1/2) (〖𝑛𝐢〗_4 )/(〖𝑛𝐢〗_(π‘›βˆ’4) ). 6^((𝑛 βˆ’ 8)/4)= 6^(1/2) Comparing powers of 6 (𝑛 βˆ’ 8)/4 = 1/2 n – 8 = 4/2 n – 8 = 2 n = 2 + 8 n = 10 Hence, n = 10

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Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 9 years. He provides courses for Maths and Science at Teachoo.