Mathematics: Prove, using a combinatorial (counting) argument, that c(2n, 2) = 2c(n, 2) + n 2

Prove, using a combinatorial (counting) argument,

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NeverEndingCycle

30.07.2019

$\begin{aligned} _{2n}c_2 = 2 ( _nc_2) + n^2 \end{aligned}$ $c(n,r) = {}_nc_r = \binom{n}{r} = \dfrac{n! }{r! (n-r)! }$ we can do $n! = n(n-1)! = n(n-1)(n-2)! = /tex]so it follows that [tex](2n)! = 2n(2n-1)! = n(2n-1)(2n-2)! = /tex]left hand side: [tex]\begin{aligned} \text{lhs} & ={} _{2n}c_2 \\ & = {} \dfrac{(2n)! }{2! (2n-2)! } \\ & = {} \dfrac{(2n)(2n-1)(2n-2)! }{2! (2n-2)! } \\ & = {} \dfrac{2n(2n-1)}{2! } \\ & = {} \dfrac{2n(2n-1)}{2} & & \text{since $2! = 2$} \\ & = {} \dfrac{n(2n-1)}{1}\\ & = {} 2n^2 - n\\ \end{aligned}$ right hand side: $\begin{aligned} \text{rhs} & = 2(_nc_2) + n^2 \\ & = 2 \left( \dfrac{n! }{2! (n-2)! } \right) + n^2 \\ & = 2 \left( \dfrac{n(n-1)(n-2)! }{2! (n-2)! } \right) + n^2 \\ & = 2 \left( \dfrac{n(n-1)}{2} \right) + n^2 \\ & = (1) \left( \dfrac{n(n-1)}{1} \right) + n^2 \\ & = n(n-1) + n^2 \\ & = n^2 - n + n^2 \\ & = 2n^2 - n \\ & = \text{lhs} \end{aligned}$

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Ответ:

JamierW2005

24.04.2020

Mathematics

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Prove, using a combinatorial (counting) argument, that c(2n, 2) = 2c(n, 2) + n 2

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