Tentukan jumlah dua deret berikut 1. 0,515151... 2. 4/3 + 4/9 + 4/81 + ....

1. kurang jelas maaf
2. 4/3,4/9,4/81
a=4/3,r=1/3
S(infinity)=a/1-r
=4/3 / 1-1/3
=4/3 / 2/3
=4/3 . 3/2=2
jadikan jawaban terbaik jika membantu
Berikan terima kasih

1) 0,515151.... = 0,51 + 0,0051 + 0,000051 + .....
a = 0,51
r = 0,0051/0,51 = 0,01
S~ = a/(1 - r) = 0,51/(1 - 0,0,1) = 0,51/0,99 = 51/99

2) seharusnya no 2 biar rasionya tetep 1/3 barisannya haruslah
4/3 + 4/9 + 4/27 + 4/81 + ...
a = 4/3
r = (4/9)/(4/3) = (4/9) . (3/4) = 3/9 = 1/3
S~ = a/(1 - r) = (4/3) / (1 - 1/3) = (4/3) / (2/3) = 4/2 = 2

CARA LAIN NO 1
Misal
a = 0,515151.... ==========> kali 100
100a = 51,515151...
--------------------------- -
-99a = -51
a = 51/99