a) tính tổng : P= 3+3^2+3^3+....+3^2023+3^2024 b) chứng tỏ rằng tổng: Q =1+3+3^2+3^3+.....+3^101 chia hết cho 13 3^2 nghĩa là ba mũ hai

`a)`

`P=3+3^2+3^3+...+3^2023+3^2024`

`=>3P=3(3+3^2+3^3+...+3^2023+3^2024)`

`=>3P=3^2+3^3+3^4+...+3^2024+3^2025`

`=>3P-P=(3^2+3^3+3^4+...+3^2024+3^2025)-(3+3^2+3^3+...+3^2023+3^2024)`

`=>2P=3^2025-3`

`=>P=[3^2025-3]/[2]`

Vậy `P=[3^2025-3]/[2]`

`b)`

`Q=1+3+3^2+3^3+...+3^101`

`=(1+3+3^2)+(3^3+3^4+3^5)+...+(3^99+3^100+3^101)`

`=1.(1+3+3^2)+3^3.(1+3+3^2)+...+3^99.(1+3+3^2)`

`=13+3^3.13+...+3^99.13`

`=13.(1+3^3+...+3^99)`

`=>Q \vdots 13`