Tính tổng P = 1 + 5^3 + 5^6 + 5^9 + + 5^99 P = d5^102
A.\(P = \dfrac{{{5^{102}} - 1}}{{124}}\)
B.\(P = \dfrac{{{5^{102}} + 1}}{{124}}\)
C.\(P = \dfrac{{{5^{102}} - 1}}{{125}}\)
D.\(P = \dfrac{{{5^{102}} + 1}}{{125}}\)
A.\(P = \dfrac{{{5^{102}} - 1}}{{124}}\)
B.\(P = \dfrac{{{5^{102}} + 1}}{{124}}\)
C.\(P = \dfrac{{{5^{102}} - 1}}{{125}}\)
D.\(P = \dfrac{{{5^{102}} + 1}}{{125}}\)
Loga
Hóa Học lớp 12
