2n

5 [(2n+1)! / (n+2)!] = 3(2n-1)! / (n-1)! maka nilai n yakni 5[(2n+1)!/(2n-1)!] = 3[(2n-1)!/(n-1)!][5(2n+1)(2n)(2n-1)!] / (2n-1)! = [3(n+2)(n+1)(n)(n-1)!]/(n-1)!5(2n+1)=3(n+2)(n+1)20n+10 = 3n^2 + 9n +63n^2-11n-4=0 n= 4 atau n= -1/3 1²+3²+5²+…..+(2n-1)²=n(2n-1)(2n+1)/2​ …

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