# Misc 1 - Chapter 8 Class 11 Binomial Theorem

Last updated at May 29, 2018 by Teachoo

Last updated at May 29, 2018 by Teachoo

Transcript

Misc 1 Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively. We know that (a + b)n = nC0 an b0 + nC1 an 1 b1 + nC2 an 2 b2 + . . + nCn 1 a1 bn 1 + nCn a0 bn = an + nC1 an 1 b1 + nC2 an 2b2 + . . + nCn 1 a1 bn 1 + bn So first 3 terms are an , nC1 an 1b and nC2 an 2 b2 Also, it is given that their value are 729,7290 and 30375 Therefore, an = 729 nC1 an 1 b = 7290 nC2 an 2 b2 = 30375 Dividing (1) by (2) 1 1 = 729 7290 ! 1! 1 ! 1 = 1 10 ( 1)! 1 1 ! 1 = 1 10 1 = 1 10 ( 1) = 1 10 1 = 1 10 = 1 10 Dividing (2) by (3) 1 1 2 2 2 = 7290 30375 ! 1! 1 ! 1 . ! 2! 2 ! 2 2 = 6 25 ! 1 1 ! 2! 2 ! ! 2 2 = 6 25 ! 2! 2 ! ! 1 ! 1 2 2 = 6 25 2 ! 2 ! ! 1 2 ! 1 2 2 = 6 25 2 1 1 ( 2) 1 2 = 6 25 2 ( 1) 1 +2 = 6 25 2 1 = 6 25 Now, our equations are = 1 10 (4) 2 1 = 6 25 (5) Dividing (4) by (5) 2 1 = 1 10 6 25 1 2 = 1 10 25 6 1 2 = 5 12 12 (n 1) = 2n (5) 12n 12 = 10n 12n 10n = 12 2n = 12 n = 12 2 n = 6 Putting n = 6 in (1) an = 729 a6 = 729 a6 = (3)6 a = 3 Putting a = 3 , n = 6 in (5) 2 1 = 6 25 3 6 = 1 10 1 2 = 1 10 10 2 = b 5 = b b = 5 Hence, a = 3 , b = 5 & n = 6

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Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 8 years. He provides courses for Maths and Science at Teachoo. You can check his NCERT Solutions from Class 6 to 12.