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Example 6 - Show that middle term of (1 + x)2n is 1.3.5...(2n-1)/n - Examples

  1. Chapter 8 Class 11 Binomial Theorem
  2. Serial order wise
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Example 6 Show that the middle term in the expansion of (1 + x)2n is īˇ1 . 3 . 5 â€Ļ. (2𝑛 − 1)īˇŽđ‘›!īˇ¯ 2n xn, where n is a positive integer. Number of terms = 2n which is even So middle term = (īˇ2nīˇŽ2īˇ¯ + 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 = īˇīˇ2𝑛īˇ¯!īˇŽđ‘›!īˇ2𝑛 −𝑛īˇ¯!īˇ¯ . (1)n . xn = īˇīˇ2𝑛īˇ¯!īˇŽđ‘›! 𝑛!īˇ¯ . xn = īˇ2𝑛īˇ2𝑛−1īˇ¯īˇ2𝑛−2īˇ¯ â€Ļâ€Ļ.4 × 3 × 2 × 1īˇŽđ‘›! 𝑛!īˇ¯ xn = īˇīˇīˇ2𝑛−1īˇ¯īˇ2𝑛−3īˇ¯â€Ļ.â€Ļ. × 5 × 3 × 1īˇ¯ [īˇ2𝑛īˇ¯īˇ2𝑛−2īˇ¯â€Ļ × 4 × 2]īˇŽđ‘›! 𝑛!īˇ¯ xn = īˇīˇ1 × 3 × 5 × â€Ļâ€Ļ × īˇ2𝑛−3īˇ¯īˇ2𝑛−1īˇ¯īˇ¯ [2 × 4 × 6â€Ļ.. × īˇ2𝑛−2īˇ¯ × 2𝑛]īˇŽđ‘›! 𝑛!īˇ¯ xn = īˇīˇ1 × 3 × 5â€Ļâ€Ļ. × īˇ2𝑛−3īˇ¯īˇ2𝑛−1īˇ¯īˇ¯ [īˇ2 × 1īˇ¯ ×(2 × 2) ×(2 × 3) × â€Ļ.. × 2 īˇđ‘›âˆ’1īˇ¯ × 2(𝑛)]īˇŽđ‘›! 𝑛!īˇ¯ xn = īˇīˇ1 × 3 × 5â€Ļâ€Ļ. × īˇ2𝑛−3īˇ¯īˇ2𝑛−1īˇ¯īˇ¯ īˇ2 × 2 × 2 × 2 â€Ļâ€Ļ..× 2īˇ¯ [1 × 2 × 3 â€Ļ.. īˇđ‘›âˆ’1īˇ¯ 𝑛]īˇŽđ‘›! 𝑛!īˇ¯ xn = īˇīˇ1 × 3 × 5â€Ļâ€Ļ. × īˇ2𝑛−3īˇ¯īˇ2𝑛−1īˇ¯īˇ¯ 2𝑛 [1 × 2 × 3 â€Ļ.. īˇđ‘›âˆ’1īˇ¯ 𝑛]īˇŽđ‘›!𝑛!īˇ¯ xn = īˇīˇ1 × 3 × 5â€Ļâ€Ļ. × īˇ2n−3īˇ¯īˇ2n−1īˇ¯īˇ¯2n (n!)īˇŽn! n!īˇ¯ . xn = īˇ1 × 3 × 5â€Ļâ€Ļ. × īˇ2n−1īˇ¯īˇŽn! īˇ¯ 2n . xn Hence middle term of expansion (1 + x)2n is īˇ1 . 3 . 5â€Ļâ€Ļ. â€Ļ.īˇ2n−1īˇ¯īˇŽn! īˇ¯ 2n . xn Hence proved

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