Chứng minh 3^x+1 +3^x+2 +3^x+2 +….+ 3^x+100 chia hết cho 120 (x thuộc N) 20/10/2021 Bởi Isabelle Chứng minh 3^x+1 +3^x+2 +3^x+2 +….+ 3^x+100 chia hết cho 120 (x thuộc N)
Đáp án: Gọi A=3^x+1 +3^x+2 +3^x+2 +….+ 3^x+100 A=(3^x+1 +3^x+2+ 3^x+3+3^x+4)+…………….+(3^x+97+3^x+98+3^x+99+3^x+100) A=(3^x.3+3^x.9+3^x.27+3^x.81)+….+(3^x+96.3+3^x+96.9+3^x+96.27+3^x+96.81) A=3^x.(3+9+27+81)+………….+(3^x+96).(3+9+27+81) A=(3^x).120+(3^x+4).120+…………..+(3^x+96) . 120 A=120.(3^x +(3^x+4) +……..+3^x+96) chia hết cho 120 Vậy A chia hết cho 120 Cho mình xin hay nhất nhé Thank you very much Bình luận
Đáp án + Giải thích các bước giải: Ta có : `3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}+….+3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}` `=(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4})+….+(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100})` `=3^{x}(3+3^{2}+3^{3}+3^{4})+….+3^{x+96}(3+3^{2}+3^{3}+3^{4})` `=3^{x}.120+…+3^{x+96}.120` `=120.(3^{x}+….+3^{x+96})` $\vdots$ `120` Vậy `3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}+….+3^{x+97}+3^{98}+3^{x+99}+3^{x+100}` $\vdots$ `120` `∀x∈N` Bình luận
Đáp án:
Gọi A=3^x+1 +3^x+2 +3^x+2 +….+ 3^x+100
A=(3^x+1 +3^x+2+ 3^x+3+3^x+4)+…………….+(3^x+97+3^x+98+3^x+99+3^x+100)
A=(3^x.3+3^x.9+3^x.27+3^x.81)+….+(3^x+96.3+3^x+96.9+3^x+96.27+3^x+96.81)
A=3^x.(3+9+27+81)+………….+(3^x+96).(3+9+27+81)
A=(3^x).120+(3^x+4).120+…………..+(3^x+96) . 120
A=120.(3^x +(3^x+4) +……..+3^x+96) chia hết cho 120
Vậy A chia hết cho 120
Cho mình xin hay nhất nhé
Thank you very much
Đáp án + Giải thích các bước giải:
Ta có :
`3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}+….+3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100}`
`=(3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4})+….+(3^{x+97}+3^{x+98}+3^{x+99}+3^{x+100})`
`=3^{x}(3+3^{2}+3^{3}+3^{4})+….+3^{x+96}(3+3^{2}+3^{3}+3^{4})`
`=3^{x}.120+…+3^{x+96}.120`
`=120.(3^{x}+….+3^{x+96})` $\vdots$ `120`
Vậy `3^{x+1}+3^{x+2}+3^{x+3}+3^{x+4}+….+3^{x+97}+3^{98}+3^{x+99}+3^{x+100}` $\vdots$ `120` `∀x∈N`