web analytics

Misc 8 - Find n, if ratio of fifth term from beginning to end - Miscellaneous

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

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 ﷐﷐﷐4﷮2﷯+ ﷐1﷮﷐4﷮3﷯﷯﷯﷮𝑛﷯ is ﷐﷮6﷯ : 1 We know that General term of expansion (a + b)n Tr+1 = nCr an – rbr We need to calculate fifth term from beginning of expansion ﷐﷐﷐4﷮2﷯+ ﷐1﷮﷐4﷮3﷯﷯﷯﷮𝑛﷯ i.e. we need to calculate T5 = T4+1 Putting r = 4 , a = ﷐4﷮2﷯ , b = ﷐﷐1﷮﷐4﷮3﷯﷯﷯ T4+1 = nC4﷐﷐4﷮2﷯﷯n – 4.﷐﷐1﷮﷐4﷮3﷯﷯﷯4 T5 = nC4 ﷐((2)﷮﷐1﷮4﷯﷯)n – 4 .﷐﷐1﷮﷐3﷮﷐1﷮4﷯﷯﷯﷯4 = nCr ﷐(2)﷮﷐1﷮4﷯ ×(𝑛−4)﷯ .﷐﷐1﷮﷐3﷮﷐1﷮4﷯ ×4﷯﷯﷯ = nCr ﷐(2)﷮﷐1﷮4﷯ ×(𝑛−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 Now we need to calculate fifth term from end Hence 5th term from end = (n – 3)th term from beginning Another method, rth term from end = (n – r + 2)th term from stating Hence, 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 (﷐4﷮2﷯ + ﷐1﷮﷐4﷮3﷯﷯ )r Tr + 1 = nCr an – rbr Putting r = (n – 3) – 1 = n – 4 a = ﷐4﷮2﷯ , b = ﷐1﷮﷐4﷮3﷯﷯ T(n – 4 + 1) = nCn – 4 ﷐﷐﷐4﷮2﷯﷯﷮𝑛 −(𝑛−4)﷯ . ﷐﷐﷐1﷮﷐4﷮3﷯﷯﷯﷮𝑛 − 4﷯ = nCn – 4 ﷐﷐﷐(2)﷮﷐1﷮4﷯﷯﷯﷮n – n + 4﷯ . ﷐﷐﷐1﷮﷐3﷮﷐1﷮4﷯﷯﷯﷯﷮𝑛 − 4﷯ = nCn – 4 ﷐﷐﷐(2)﷮﷐1﷮4﷯﷯﷯﷮4﷯ . ﷐﷐﷐3﷮−﷐1﷮4﷯﷯﷯﷮𝑛 − 4﷯ = nCn – 4 ﷐(2)﷮﷐4﷮4﷯﷯ .﷐3﷮﷐−(𝑛 − 4)﷮4﷯﷯ = nCn – 4 ﷐(2)﷮1﷯ .﷐﷐﷐1﷮3﷯﷯﷮﷐𝑛 − 4﷮4﷯﷯ Given that ratio fifth term from beginning fifth term from end is ﷐﷮6﷯ : 1 ﷐𝐹𝑖𝑓𝑡ℎ 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑏𝑒𝑔𝑖𝑛𝑛𝑖𝑛𝑔 ﷮𝐹𝑖𝑓𝑡ℎ 𝑡𝑒𝑟𝑚 𝑓𝑟𝑜𝑚 𝑒𝑛𝑑 ﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑇﷮5﷯﷮﷐𝑇﷮𝑛−3﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ Putting values from (2) & (3) ﷐﷐𝑛𝐶﷮4﷯ ﷐2﷮﷐𝑛−4﷮4﷯﷯ . ﷐1﷮3﷯﷮﷐𝑛𝐶﷮𝑛−4﷯ 2 . ﷐﷐﷐1﷮3﷯﷯﷮﷐𝑛−4﷮4﷯﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷐2﷮﷐𝑛−4﷮4﷯﷯ . ﷐1﷮3﷯﷮﷐𝑛𝐶﷮𝑛−4﷯ 2 . ﷐﷐﷐1﷮3﷯﷯﷮﷐𝑛−4﷮4﷯﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯ . ﷐﷐2﷯﷮﷐𝑛−4﷮4﷯ −1 ﷯﷐﷐﷐1﷮3﷯﷯﷮ 1−﷐﷐𝑛−4﷮4﷯﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯ . ﷐2﷮﷐𝑛−4−4﷮4﷯ ﷯. ﷐﷐﷐1﷮3﷯﷯﷮﷐4−𝑛+4﷮4﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯. ﷐2﷮﷐𝑛−8﷮4﷯ ﷯. ﷐﷐﷐1﷮3﷯﷯﷮﷐8−𝑛﷮4﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯. ﷐2﷮﷐𝑛−8﷮4﷯ ﷯. ﷐﷐﷐1﷮3﷯﷯﷮﷐−(𝑛−8)﷮4﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯. ﷐2﷮﷐𝑛−8﷮4﷯﷯. ﷐﷐3﷯﷮﷐𝑛−8﷮4﷯﷯ = ﷐﷐﷮6﷯﷮1﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯. ﷐(2 ×3)﷮﷐𝑛−8﷮4﷯﷯. = ﷐6﷮﷐1﷮2﷯﷯ ﷐﷐𝑛𝐶﷮4﷯ ﷮﷐𝑛𝐶﷮𝑛−4﷯ ﷯. ﷐6﷮﷐𝑛−8﷮4﷯﷯= ﷐6﷮﷐1﷮2﷯﷯ Comparing powers of 6 ﷐𝑛 − 8﷮4﷯ = ﷐1﷮2﷯ n – 8 = ﷐4﷮2﷯ n – 8 = 2 n = 2 + 8 n = 10 Hence n = 10

About the Author

Davneet Singh's photo - Teacher, Computer Engineer, Marketer
Davneet Singh
Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.
Jail