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 +6
3n^2-11n-4=0
n= 4 atau n= -1/3
1²+3²+5²+…..+(2n-1)²=n(2n-1)(2n+1)/2
Jawab:
Penjelasan dgn tindakan:
Buktikan bahwa 1^2+3^2+5^2+….+(2n+1)^2=1/3 (n+1)(2n+1)(2n+3), n>0
Penjelasan dgn langkah-langkah:
agar berfaedah.
maaf kalo salah
5 x (2n+1)!/(n+2)! = 3 x (2n-1)!/(n-1)!
Jawab:
notasi faktorial
n! = n . (n -1) .(n- 2)!
Penjelasan dgn langkah-langkah:
[ 5 (2n + 1) ! ] / (n + 2)! = [ 3 (2n – 1 ) ! / (n – 1 ) ! ]
5(2n + 1)! / (2n- 1)! = 3 ( n+2)! / (n – 1)!
5(2n + 1)(2n) = 3 (n +2)(n + 1) (n)
10 (2n + 1) = 3 (n² + 3n + 2)
20 n + 10 = 3n² + 9n + 6
.
3n² +9n + 6 – 20n – 10 =0
3n² – 11n – 4 = 0
(3n + 1)(n – 4 ) =0
n = – 1/3 atau n = 4
syarat i) (n – 2) ! –> n > 2 , ii ) (n – 1) ! –> n > 1
HP n = – 1/3 atau n = 4 dgn x > 2 dan n> 1
HP n yg memenuhi n = 4
5 [(2n+1)! / (n+2)!] = 3(2n-1)! / (n-1)! maka nilai n adalah
maaf kalo salah ya, semangat ![5 [(2n+1)! / (n+2)!] = 3(2n-1)! / (n-1)! maka nilai n adalah](https://id-static.z-dn.net/files/d3a/ca83965cb0e17a8a9398b652fe827406.jpg "5 [(2n+1)! / (n+2)!] = 3(2n-1)! / (n-1)! maka nilai n adalah")