549 people joined Teachoo Black. What are you waiting for?

Miscellaneous

Misc 1
Important
Deleted for CBSE Board 2022 Exams

Misc 2 Deleted for CBSE Board 2022 Exams

Misc 3 Important Deleted for CBSE Board 2022 Exams

Misc 4 Deleted for CBSE Board 2022 Exams

Misc 5 Important Deleted for CBSE Board 2022 Exams

Misc 6 Deleted for CBSE Board 2022 Exams

Misc 7 Important Deleted for CBSE Board 2022 Exams

Misc 8 Important Deleted for CBSE Board 2022 Exams You are here

Misc 9 Important Deleted for CBSE Board 2022 Exams

Misc 10 Deleted for CBSE Board 2022 Exams

Chapter 8 Class 11 Binomial Theorem (Deleted)

Serial order wise

Last updated at Jan. 29, 2020 by Teachoo

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 (β2+ 1/β3)^π is β6 : 1 We know that General term of expansion (a + b)n Tr + 1 = nCr an β rbr Fifth term from beginning We need to calculate T5 = T4 + 1 Putting r = 4 , a = β2 , b = (1/β3) T4 + 1 = nC4(β2)n β 4.(1/β3)4 T5 = nC4 γ((2)γ^(1/4))n β 4 .(1/3^(1/4) )4 T5 = nC4 γ(2)γ^(1/4 Γ (π β 4)) .(1/3^(1/4) )4 = nCr γ(2)γ^(1/4 Γ(πβ4)) .(1/3) Finding 5th term from end We know that (a + b)n = nCo an bo +nC1 anβ1 b1 +β¦β¦..+ nCnβ1 (a)n β (nβ1) .bnβ1 +nCn a0 bn = an + nC1 anβ1b1 + β¦β¦β¦β¦β¦β¦β¦β¦β¦+ nC1 a1bnβ1 + bn = bn + nC1 a1 bnβ1 +β¦β¦β¦β¦β¦β¦β¦+ nC1 anβ1 b1 + an Hence, 5th term from end = (n β 3)th term from beginning Another method, rth term from end = (n β r + 2)th term from stating 5th term from end = (n β 5 + 2)th term from stating = (n β 3)th term from beginning Now we need to calculate (n β 3)th term of expansion (β(π&π) + π/β(π&π) )r Tr + 1 = nCr an β rbr Putting r = (n β 3) β 1 = n β 4 a = β2 , b = 1/β3 T(n β 4 + 1) = nCn β 4 (β2)^(π β(πβ4)) . (1/β3)^(π β 4) = nCn β 4 [γ(2)γ^(1/4) ]^"n β n + 4" . (1/3^(1/4) )^(π β 4) = nCn β 4 [γ(2)γ^(1/4) ]^"4" . (3^((β1)/4) )^(π β 4) = nCn β 4 γ(2)γ^1 .(1/3)^((π β 4)/4) = nCn β 4 γ(2)γ^(4/4) .3^((β(π β 4))/4) = nCn β 4 γ(2)γ^1 .(1/3)^((π β 4)/4) Given that Ratio fifth term from beginning fifth term from end is β6 : 1 (πΉπππ‘β π‘πππ ππππ πππππππππ )/(πΉπππ‘β π‘πππ ππππ πππ ) = β6/1 (γππΆγ_4 2^((πβ4)/4) . 1/3)/(γππΆγ_(πβ4) 2 . (1/3)^((πβ4)/4) ) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ) . (2)^((π β 4)/4 β1 ) (1/3)^( 1 β ((π β 4)/4) ) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ) . 2^((π β 4 β 4)/4 ). (1/3)^((4 β π + 4)/4) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ). 2^((π β 8)/4 ). (1/3)^((8 β π)/4) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ). 2^((π β 8)/4 ). (1/3)^((β(π β 8))/4) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ). 2^((πβ8)/4). (3)^((π β 8)/4) = β6/1 (γππΆγ_4 )/(γππΆγ_(πβ4) ). γ(2 Γ 3)γ^((π β 8)/4). = 6^(1/2) (γππΆγ_4 )/(γππΆγ_(πβ4) ). 6^((π β 8)/4)= 6^(1/2) Comparing powers of 6 (π β 8)/4 = 1/2 n β 8 = 4/2 n β 8 = 2 n = 2 + 8 n = 10 Hence, n = 10