2.tentukan empat suku pertama suatu barisan yang suku ke n nya dinyatakan dengan rumus berikut
a. 4n+5
b. 5x2n

Jawaban:
4,8,16,32,...
a = 4
r = 2
U9 = a.r^n-1
= 4. 2^9-1
= 4. 2^8
= 4. 256
= 1024
4n+5
n = 0 ---> 4 (0) + 5 = 5
n = 1 ---> 4 (1) + 5 = 9
n = 2 ---> 4 (2) + 5 = 13
n = 3 ---> 4 (3) + 5 = 15
4 barisan pertama : 5,9,13,15
5x2n
n = 0 ---> 5 x 2(0) = 5 x 0 = 0
n = 1 ---> 5 x 2(1) = 5 x 2 = 10
n = 2 ---> 5 x 2(2) = 5 x 4 = 20
n = 3 ---> 5 x 2(3) = 5 x 6 = 30
4 barisan pertama : 0,10,20,30
Jawab:
1. 1024
Penjelasan dengan langkah-langkah:
1.
4,8,16,32,64,128,256,512,1024
2.
4n+5
n = 0 > 4 (0) + 5 = 5
= 1 > 4 (1) + 5 = 9
= 2 > 4 (2) + 5 = 13
= 3 > 4 (3) + 5 = 15
4 barisan pertama : 5,9,13,15
5x2n
n = 0 > 5 x 2(0) = 5 x 0 = 0
= 1 > 5 x 2(1) = 5 x 2 = 10
= 2 > 5 x 2(2) = 5 x 4 = 20
= 3 > 5 x 2(3) = 5 x 6 = 30
4 barisan pertama : 0,10,20,30
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