Ta có: 2n2−n+22n+1=2n2+n−2n−1+32n+12n2−n+22n+1=2n2+n−2n−1+32n+1
= n(2n+1)−(2n+1)+32n+1n(2n+1)−(2n+1)+32n+1
=(2n+1)(n−1)+32n+1=n−1+32n+1=(2n+1)(n−1)+32n+1=n−1+32n+1
Để 2n2−n+22n2−n+2 chia hết cho 2n+12n+1 (với n∈Z)n∈Z) thì 2n+12n+1 phải là ước của 33, hay 2n+1∈{1;−1;3;−3}2n+1∈{1;−1;3;−3}.
+) 2n+1=1⇒2n=0⇒n=02n+1=1⇒2n=0⇒n=0
+) 2n+1=−1⇒2n=−2⇒n=−12n+1=−1⇒2n=−2⇒n=−1
+) 2n+1=3⇒2n=2⇒n=12n+1=3⇒2n=2⇒n=1
+) 2n+1=−3⇒2n=−4⇒n=−22n+1=−3⇒2n=−4⇒n=−2
Vậy n∈{0;−2;−1;1}n∈{0;−2;−1;1}