Cho T=3+3^2+...+3^99. a) Tìm số tự nhiên n để 2T + 3 = 3^2n b) CMR T chia hết cho 13

`a) T=3+3^2+3^3+.......+3^99`

`3T=3^2+3^3+3^4.......+3^100`

`3T-T=(3^2+3^3+3^4.......+3^100)-(3+3^2+3^3+.......+3^99)`

`2T= 3^100-3`

`⇒` Ta có : `2T+3=3^(2n)`

`⇔ 3^100-3+3=3^(2n)`

`⇔ 3^100=3^(2n)`

`⇔ 2n=100`

`⇔ n=100:2`

`⇔ n=50`

`b) T=3+3^2+3^3+.......+3^99`

`= (3+3^2+3^3)+(3^4+3^5+3^6).......+(3^97+3^98+3^99)`

`= (3+3^2+3^3)+3^3.(3+3^2+3^3)+........+3^96.(3+3^2+3^3)`

`= 39+3^3.39+.......3^96.39`

`=39.(1+3^3+.........+3^96)`

Vì `39 \vdots 13 ⇒ 39.(1+3^3+.........+3^96) \vdots 13` ( điều phải chứng minh)

$\text{b, T =3+3$^{2}$ +...+3$^{99}$}$

$\text{T = (3 + 3$^{2}$ + 3$^{3}$) + ... + (3$^{97}$ + 3$^{98}$ + 3$^{99}$)}$
$\text{T = (3 + 3$^{2}$ + 3$^{3}$) + ... + 3$^{97}$.(3 + 3$^{2}$ + 3$^{3}$)}$
$\text{T = 39 + ... + 3$^{97}$.39}$
$\text{Vì 39 ⋮ 13 nên T ⋮ 13}$
$\text{Vậy T ⋮ 13}$ $\textit{(đpcm)}$