tìm n biết : $3^{n+3}$ + $2^{n+3}$ – $3^{n+2}$ + $2^{n+2}$ chia hết cho 6 17/07/2021 Bởi Faith tìm n biết : $3^{n+3}$ + $2^{n+3}$ – $3^{n+2}$ + $2^{n+2}$ chia hết cho 6
Ta có: $3^{n+3}+2^{n+3}-3^{n+2}+2^{n+2}$ $=27.3^n+8.2^n-9.3^n+4.2^n$$=3^n.(27-9)+2^n.(8+14)$ $=3^n.18+2^n.12$$=6.(3^n.3+2^n.2)$ $=6.(3^{n+1}+2^{n+1})$ Vậy: n∈R Bình luận
$3^{n+3}+2^{n+3}-3^{n+2}+2^{n+2}$ = $3^{n+2}.(3-1)+2^{n+2}.( 2+1)$= $3^{n+2}.2+2^{n+2}.3$= $3^{n+1}.3.2+2^{n+1}.2.3$= $3^{n+1}.6+2^{n+1}.6$= $6.( 3^{n+1}+2^{n+1})$ ⇒ $3^{n+1}+2^{n+1}$ là số nguyên ⇒ $n+1≥ 0$ ⇔ $n≥ -1$ Bình luận
Ta có:
$3^{n+3}+2^{n+3}-3^{n+2}+2^{n+2}$
$=27.3^n+8.2^n-9.3^n+4.2^n$$=3^n.(27-9)+2^n.(8+14)$
$=3^n.18+2^n.12$$=6.(3^n.3+2^n.2)$
$=6.(3^{n+1}+2^{n+1})$
Vậy: n∈R
$3^{n+3}+2^{n+3}-3^{n+2}+2^{n+2}$
= $3^{n+2}.(3-1)+2^{n+2}.( 2+1)$
= $3^{n+2}.2+2^{n+2}.3$
= $3^{n+1}.3.2+2^{n+1}.2.3$
= $3^{n+1}.6+2^{n+1}.6$
= $6.( 3^{n+1}+2^{n+1})$
⇒ $3^{n+1}+2^{n+1}$ là số nguyên
⇒ $n+1≥ 0$
⇔ $n≥ -1$