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`
