Given a sequence 0,65,365,21635,..., the 12th term of this sequence is....
Explanation
Given sequence
0,65,365,21635,...
Each term if expanded becomes
u1=0=201−301=21−11+(−1)1⋅3(1−1)1
u2=65=211+311=22−11+(−1)2⋅3(2−1)1
u3=365=221−321=23−11+(−1)3⋅3(3−1)1
u4=21635=231+331=24−11+(−1)4⋅3(4−1)1
Therefore, the nth term can be formulated as follows
un=2n−11+(−1)n⋅3(n−1)1
Then the 12th term is
u12=212−11+(−1)12⋅3(12−1)1
u12=2111+1(3(11)1)
u12=2111+3111