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

Transcript

Example 6 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 โ€ฆ. (2๐‘› โˆ’ 1))/๐‘›! 2n xn, where n is a positive integer. Given Number of terms = 2n which is even So, Middle term = (2n/2 + 1)th term = (n + 1)th term Hence, we need to find Tn + 1 We know that general term of (a + b)nis Tr + 1 = nCr an โ€“ r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n โ€“ n (x)n = (2๐‘›)!/๐‘›!(2๐‘› โˆ’๐‘›)! . (1)n . xn = (2๐‘›)!/(๐‘›! ๐‘›!) . xn = (2๐‘›(2๐‘› โˆ’ 1)(2๐‘› โˆ’ 2) โ€ฆโ€ฆ. 4 ร— 3 ร— 2 ร— 1)/(๐‘›! ๐‘›!) xn = ([(2๐‘› โˆ’ 1)(2๐‘› โˆ’ 3)โ€ฆ.โ€ฆ. ร— 5 ร— 3 ร— 1] [(2๐‘›)(2๐‘› โˆ’ 2)โ€ฆ ร— 4 ร— 2])/(๐‘›! ๐‘›!) xn We need to show (1 . 3 . 5 โ€ฆ. (2๐‘› โˆ’ 1))/๐‘›! 2n xn = ([1 ร— 3 ร— 5 ร— โ€ฆโ€ฆ ร— (2๐‘› โˆ’ 3)(2๐‘› โˆ’ 1)] [2 ร— 4 ร— 6โ€ฆ.. ร— (2๐‘› โˆ’ 2) ร— 2๐‘›])/(๐‘›! ๐‘›!) xn = ([1 ร— 3 ร— 5โ€ฆโ€ฆ. ร— (2๐‘›โˆ’3)(2๐‘›โˆ’1)] [(2 ร— 1) ร—(2 ร— 2) ร—(2 ร— 3) ร— โ€ฆ.. ร— 2 (๐‘›โˆ’1) ร— 2(๐‘›)])/(๐‘›! ๐‘›!) xn = ([1 ร— 3 ร— 5โ€ฆโ€ฆ. ร— (2๐‘›โˆ’3)(2๐‘›โˆ’1)] (2 ร— 2 ร— 2 ร— 2 โ€ฆโ€ฆ..ร— 2) [1 ร— 2 ร— 3 โ€ฆ.. (๐‘›โˆ’1) ๐‘›])/(๐‘›! ๐‘›!) xn = ([1 ร— 3 ร— 5โ€ฆโ€ฆ. ร— (2๐‘› โˆ’ 3)(2๐‘› โˆ’ 1)] 2๐‘› [1 ร— 2 ร— 3 โ€ฆ.. (๐‘› โˆ’ 1) ๐‘›])/๐‘›!๐‘›! xn = ([1 ร— 3 ร— 5โ€ฆโ€ฆ. ร— (2n โˆ’ 3)(2n โˆ’ 1)] 2n (n!))/(n! n!) . xn = (๐Ÿ ร— ๐Ÿ‘ ร— ๐Ÿ“โ€ฆโ€ฆ. ร— (๐Ÿ๐ง โˆ’ ๐Ÿ))/(๐ง! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5โ€ฆโ€ฆ. โ€ฆ.(2nโˆ’1))/(n! ) 2n . xn Hence proved

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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.