# Example 6

Last updated at Dec. 8, 2016 by Teachoo

Last updated at Dec. 8, 2016 by Teachoo

Transcript

Example 6 Show that the middle term in the expansion of (1 + x)2n is īˇ1 . 3 . 5 âĻ. (2đ â 1)īˇŽđ!īˇ¯ 2n xn, where n is a positive integer. Number of terms = 2n which is even So middle term = (īˇ2nīˇŽ2īˇ¯ + 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)2nân (x)n = īˇīˇ2đīˇ¯!īˇŽđ!īˇ2đ âđīˇ¯!īˇ¯ . (1)n . xn = īˇīˇ2đīˇ¯!īˇŽđ! đ!īˇ¯ . xn = īˇ2đīˇ2đâ1īˇ¯īˇ2đâ2īˇ¯ âĻâĻ.4 Ã 3 Ã 2 Ã 1īˇŽđ! đ!īˇ¯ xn = īˇīˇīˇ2đâ1īˇ¯īˇ2đâ3īˇ¯âĻ.âĻ. Ã 5 Ã 3 Ã 1īˇ¯ [īˇ2đīˇ¯īˇ2đâ2īˇ¯âĻ Ã 4 Ã 2]īˇŽđ! đ!īˇ¯ xn = īˇīˇ1 Ã 3 Ã 5 Ã âĻâĻ Ã īˇ2đâ3īˇ¯īˇ2đâ1īˇ¯īˇ¯ [2 Ã 4 Ã 6âĻ.. Ã īˇ2đâ2īˇ¯ Ã 2đ]īˇŽđ! đ!īˇ¯ xn = īˇīˇ1 Ã 3 Ã 5âĻâĻ. Ã īˇ2đâ3īˇ¯īˇ2đâ1īˇ¯īˇ¯ [īˇ2 Ã 1īˇ¯ Ã(2 Ã 2) Ã(2 Ã 3) Ã âĻ.. Ã 2 īˇđâ1īˇ¯ Ã 2(đ)]īˇŽđ! đ!īˇ¯ xn = īˇīˇ1 Ã 3 Ã 5âĻâĻ. Ã īˇ2đâ3īˇ¯īˇ2đâ1īˇ¯īˇ¯ īˇ2 Ã 2 Ã 2 Ã 2 âĻâĻ..Ã 2īˇ¯ [1 Ã 2 Ã 3 âĻ.. īˇđâ1īˇ¯ đ]īˇŽđ! đ!īˇ¯ xn = īˇīˇ1 Ã 3 Ã 5âĻâĻ. Ã īˇ2đâ3īˇ¯īˇ2đâ1īˇ¯īˇ¯ 2đ [1 Ã 2 Ã 3 âĻ.. īˇđâ1īˇ¯ đ]īˇŽđ!đ!īˇ¯ xn = īˇīˇ1 Ã 3 Ã 5âĻâĻ. Ã īˇ2nâ3īˇ¯īˇ2nâ1īˇ¯īˇ¯2n (n!)īˇŽn! n!īˇ¯ . xn = īˇ1 Ã 3 Ã 5âĻâĻ. Ã īˇ2nâ1īˇ¯īˇŽn! īˇ¯ 2n . xn Hence middle term of expansion (1 + x)2n is īˇ1 . 3 . 5âĻâĻ. âĻ.īˇ2nâ1īˇ¯īˇŽn! īˇ¯ 2n . xn Hence proved

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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.