prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1) n-1 2n(2n-1)...(n+2)/(n-1)!] = (-1) n+1 (2n)!/1(n!) 2

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Aman Bansal · Answer

dear student,

applying binomial theorwm in the given question,

1) put n= 1

2) put n=k

3) put n= k+1

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prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1) n-1 2n(2n-1)...(n+2)/(n-1)!] = (-1) n+1 (2n)!/1(n!) 2

prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1)^n-12n(2n-1)...(n+2)/(n-1)!] = (-1)ⁿ⁺¹(2n)!/1(n!)²

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prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1) n-1 2n(2n-1)...(n+2)/(n-1)!] = (-1) n+1 (2n)!/1(n!) 2

prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1)n-12n(2n-1)...(n+2)/(n-1)!] = (-1)n+1(2n)!/1(n!)2

1 Answers

Think You Can Provide A Better Answer ?

Other Related Questions on Algebra

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Complete Self Study Package designed by Industry Leading Experts

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Register Now & Win Upto 25% Scholorship

prove that [1-2n +2n(n-1)/2!] - [2n(2n-1)(2n-2)/3!]...+[(-1)^n-12n(2n-1)...(n+2)/(n-1)!] = (-1)ⁿ⁺¹(2n)!/1(n!)²