Bài 4 (1,0 điểm). Tính: a) $\frac{4^{6}.9^{5} + 6^{9}.120}{8^{4}.3^{12} - 6^{11}}$ b) B = 1 + $\frac{1}{13}$ + $\frac{1}{13^{2}}$ + $\frac{1}{13^{3}}$ + ... + $\frac{1}{13^{N}}$ (n ∈ N*)

`a)``(4^6 . 9^5 +6^9 . 120)/(8^4 . 3^12 -6^11)`

`=(2^12 . 3^10 +2^9 . 3^9 . 2^3 .3.5)/(2^12 . 3^12 -2^11 . 3^11)`

`=(2^12 . 3^10 +2^12 . 3^10 . 5)/(2^12 . 3^12 - 2^11 . 3^11)`

`=(2^12 . 3^10 (1+5))/(2^11 . 3^11 (2+3-1))`

`=(2.3.6)/(4)=36/4=9`

`b)``B=1+1/13+1/(13^2)+...+1/(13^n)`

`13B=13+1+1/13+1/(13^2)+...+1/(13^(n-1)`

`13B-B=(13+1+1/13+1/(13^2)+...+1/(13^(n-1))`

`-(1+1/13+1/(13^2)+...+1/(13^n))`

`12B=13`

`<=>B=13/12`

` Vậy    B=13/12`