質問: cho a=3 mũ 1+3 mũ 2+ 3 mu 3+ 3 mũ 4+...+3 mũ 2010 chia hết cho4vaf 13 câu hỏi 1385943 - hoidap247.com cho a=3 mũ 1+3 mũ 2+ 3 mu 3+ 3 mũ 4+...+3 mũ 2010 chia hết cho4vaf 13
回答: giải thích các bước giải: ta có:`a=(3^1+3^2+3^3)+(3^4+3^5+3^6)+...+(3^2018+3^2019+3^2020)``a=39+(3^3 .3+3^3 .3^2+3^3 .3^3)+...+(3^2017 .3+3^2017 .3^2+3^2017 .3^3)``a=39+3^3 .(3+3^2+3^3)+...+3^2017.(3+3^2+3^3)``a=39.1+3^3 . 39+...+3^2017 .39``a=39.(1+3^3+...+3^2017)`vì `39\vdots13``=>a\vdots13(đpcm)` ta có:`a=(3+3^2)+(3^3+3^4)+...+(3^2019+3^2020)``a=12.1+(3^2 .3+3^2 .3^2)+...+(3^2018 .3+3^2018 .3^2)``a=12.1+3^2 .(3+3^2)+...+3^2018.(3+3^2)``a=12.1+3^2 .12+...+3^2018 .12``a=12.(1+3^2 +...+3^2018)`vì `12\vdots4``=>a\vdots4(đpcm)`