Example 6 - Chapter 8 Class 11 Binomial Theorem (Deleted)
Last updated at Jan. 29, 2020 by Teachoo
Last updated at Jan. 29, 2020 by Teachoo
Transcript
Example 6 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 โฆ. (2๐ โ 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 โ๐)! . (1)n . xn = (2๐)!/(๐! ๐!) . xn = (2๐(2๐ โ 1)(2๐ โ 2) โฆโฆ. 4 ร 3 ร 2 ร 1)/(๐! ๐!) xn = ([(2๐ โ 1)(2๐ โ 3)โฆ.โฆ. ร 5 ร 3 ร 1] [(2๐)(2๐ โ 2)โฆ ร 4 ร 2])/(๐! ๐!) xn We need to show (1 . 3 . 5 โฆ. (2๐ โ 1))/๐! 2n xn = ([1 ร 3 ร 5 ร โฆโฆ ร (2๐ โ 3)(2๐ โ 1)] [2 ร 4 ร 6โฆ.. ร (2๐ โ 2) ร 2๐])/(๐! ๐!) xn = ([1 ร 3 ร 5โฆโฆ. ร (2๐โ3)(2๐โ1)] [(2 ร 1) ร(2 ร 2) ร(2 ร 3) ร โฆ.. ร 2 (๐โ1) ร 2(๐)])/(๐! ๐!) xn = ([1 ร 3 ร 5โฆโฆ. ร (2๐โ3)(2๐โ1)] (2 ร 2 ร 2 ร 2 โฆโฆ..ร 2) [1 ร 2 ร 3 โฆ.. (๐โ1) ๐])/(๐! ๐!) xn = ([1 ร 3 ร 5โฆโฆ. ร (2๐ โ 3)(2๐ โ 1)] 2๐ [1 ร 2 ร 3 โฆ.. (๐ โ 1) ๐])/๐!๐! xn = ([1 ร 3 ร 5โฆโฆ. ร (2n โ 3)(2n โ 1)] 2n (n!))/(n! n!) . xn = (๐ ร ๐ ร ๐โฆโฆ. ร (๐๐ง โ ๐))/(๐ง! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5โฆโฆ. โฆ.(2nโ1))/(n! ) 2n . xn Hence proved
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