Get Real time Doubt solving from 8pm to 12 am!

Examples

Example 1

Example 2 Important

Example 3 Important

Example 4

Example 5 Important Deleted for CBSE Board 2023 Exams

Example 6 Important Deleted for CBSE Board 2023 Exams You are here

Example 7 Deleted for CBSE Board 2023 Exams

Example 8 Important Deleted for CBSE Board 2023 Exams

Example 9 Deleted for CBSE Board 2023 Exams

Example 10 Important Deleted for CBSE Board 2023 Exams

Example 11 Important Deleted for CBSE Board 2023 Exams

Example 12 Deleted for CBSE Board 2023 Exams

Example 13 Important Deleted for CBSE Board 2023 Exams

Example 14 Important Deleted for CBSE Board 2023 Exams

Example 15 Important Deleted for CBSE Board 2023 Exams

Example 16 Deleted for CBSE Board 2023 Exams

Example 17 Important Deleted for CBSE Board 2023 Exams

Last updated at Jan. 29, 2020 by Teachoo

Example 6 Show that the middle term in the expansion of (1 + x)2n is (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 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 – r br For Tn + 1 , Putting n = 2n , r = n , a = 1 & b = x Tn+1 = 2nCn (1)2n – n (x)n = (2𝑛)!/𝑛!(2𝑛 −𝑛)! . (1)n . xn = (2𝑛)!/(𝑛! 𝑛!) . xn = (2𝑛(2𝑛 − 1)(2𝑛 − 2) ……. 4 × 3 × 2 × 1)/(𝑛! 𝑛!) xn = ([(2𝑛 − 1)(2𝑛 − 3)….…. × 5 × 3 × 1] [(2𝑛)(2𝑛 − 2)… × 4 × 2])/(𝑛! 𝑛!) xn We need to show (1 . 3 . 5 …. (2𝑛 − 1))/𝑛! 2n xn = ([1 × 3 × 5 × …… × (2𝑛 − 3)(2𝑛 − 1)] [2 × 4 × 6….. × (2𝑛 − 2) × 2𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] [(2 × 1) ×(2 × 2) ×(2 × 3) × ….. × 2 (𝑛−1) × 2(𝑛)])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛−3)(2𝑛−1)] (2 × 2 × 2 × 2 ……..× 2) [1 × 2 × 3 ….. (𝑛−1) 𝑛])/(𝑛! 𝑛!) xn = ([1 × 3 × 5……. × (2𝑛 − 3)(2𝑛 − 1)] 2𝑛 [1 × 2 × 3 ….. (𝑛 − 1) 𝑛])/𝑛!𝑛! xn = ([1 × 3 × 5……. × (2n − 3)(2n − 1)] 2n (n!))/(n! n!) . xn = (𝟏 × 𝟑 × 𝟓……. × (𝟐𝐧 − 𝟏))/(𝐧! ) 2n . xn Hence middle term of expansion (1 + x)2n is (1 . 3 . 5……. ….(2n−1))/(n! ) 2n . xn Hence proved