prove that
(n+1)(n+2)(n+3).......2n=(2n)!÷n!
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RHS-
2n!÷n!= { [1×2×3×.........×(n-1)×n]×[(n+1)×(n+2)×......×2n] }÷ [1×2×3×.........×n]
=(n+1)×(n+2)×(n+3)×.....×2n
=LHS
Hence proved
2n!÷n!= { [1×2×3×.........×(n-1)×n]×[(n+1)×(n+2)×......×2n] }÷ [1×2×3×.........×n]
=(n+1)×(n+2)×(n+3)×.....×2n
=LHS
Hence proved
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