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) Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xot in theexpansion of (1 + x)2n-1 |
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Answer» Use the general term in the expansion of (x + y)ⁿ is T_{r+1} = nCr.x^(n-r)y^r , then we have to find require coefficients by taking the suitable value of r . proof:For, (1 + x)²ⁿ , the general term is T_{r+1} = 2nCr.x^r for coefficient of xⁿ , put r = n T_{n+1} = 2nCn.x^nso, coefficient of xⁿ = 2nCn______(1) For , (1+x)^(2n-1) , the General term is T_{r+1} = (2n-1)Cr.x^r for coefficient of xⁿ , put r = nT_{n+1} = (2n-1)Cn.x^nso, coefficient of xⁿ = (2n-1)Cn= (2n-1)!/n!×(2n-1-n)! = (2n-1)!/n! × (n-1)! multiplying and dividing by 2n = (2n)(2n-1)!/(2n)n!(n-1)!= (2n)!/2n!{n(n-1)!}= (2n)!/2n!n!= 1/2 {2nCn}__________(2) from equations (1) and (2), coefficient of xⁿ in (1 + x)^(2n-1) = 1/2 × coefficient of xⁿ in (1 + x)²ⁿ 2 × coefficient of xⁿ in (1 + x)^(2n-1) = coefficient of xⁿ in (1 + x)²ⁿ |
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