0000:00:00:00:00Start8Number 8ExplanationGiven a sequence 0,56,536,35216,...0, \frac{5}{6}, \frac{5}{36}, \frac{35}{216}, ...0,65,365,21635,..., the 121212th term of this sequence is....1211−1311\frac{1}{2^{11}} - \frac{1}{3^{11}}2111−31111211−2311\frac{1}{2^{11}} - \frac{2}{3^{11}}2111−31123211−1311\frac{3}{2^{11}} - \frac{1}{3^{11}}2113−31111211+1311\frac{1}{2^{11}} + \frac{1}{3^{11}}2111+31112211+3311\frac{2}{2^{11}} + \frac{3}{3^{11}}2112+3113ExplanationGiven sequence 0,56,536,35216,...0, \frac{5}{6}, \frac{5}{36}, \frac{35}{216}, ...0,65,365,21635,... Each term if expanded becomes u1=0=120−130=121−1+(−1)1⋅13(1−1)u_1 = 0 = \frac{1}{2^0} - \frac{1}{3^0} = \frac{1}{2^{1-1}} + (-1)^1 \cdot \frac{1}{3^{(1-1)}}u1=0=201−301=21−11+(−1)1⋅3(1−1)1u2=56=121+131=122−1+(−1)2⋅13(2−1)u_2 = \frac{5}{6} = \frac{1}{2^1} + \frac{1}{3^1} = \frac{1}{2^{2-1}} + (-1)^2 \cdot \frac{1}{3^{(2-1)}}u2=65=211+311=22−11+(−1)2⋅3(2−1)1u3=536=122−132=123−1+(−1)3⋅13(3−1)u_3 = \frac{5}{36} = \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{2^{3-1}} + (-1)^3 \cdot \frac{1}{3^{(3-1)}}u3=365=221−321=23−11+(−1)3⋅3(3−1)1u4=35216=123+133=124−1+(−1)4⋅13(4−1)u_4 = \frac{35}{216} = \frac{1}{2^3} + \frac{1}{3^3} = \frac{1}{2^{4-1}} + (-1)^4 \cdot \frac{1}{3^{(4-1)}}u4=21635=231+331=24−11+(−1)4⋅3(4−1)1 Therefore, the nnnth term can be formulated as follows un=12n−1+(−1)n⋅13(n−1)u_n = \frac{1}{2^{n-1}} + (-1)^n \cdot \frac{1}{3^{(n-1)}}un=2n−11+(−1)n⋅3(n−1)1 Then the 121212th term is u12=1212−1+(−1)12⋅13(12−1)u_{12} = \frac{1}{2^{12-1}} + (-1)^{12} \cdot \frac{1}{3^{(12-1)}}u12=212−11+(−1)12⋅3(12−1)1u12=1211+1(13(11))u_{12} = \frac{1}{2^{11}} + 1 \left(\frac{1}{3^{(11)}}\right)u12=2111+1(3(11)1)u12=1211+1311u_{12} = \frac{1}{2^{11}} + \frac{1}{3^{11}}u12=2111+3111
8Number 8ExplanationGiven a sequence 0,56,536,35216,...0, \frac{5}{6}, \frac{5}{36}, \frac{35}{216}, ...0,65,365,21635,..., the 121212th term of this sequence is....1211−1311\frac{1}{2^{11}} - \frac{1}{3^{11}}2111−31111211−2311\frac{1}{2^{11}} - \frac{2}{3^{11}}2111−31123211−1311\frac{3}{2^{11}} - \frac{1}{3^{11}}2113−31111211+1311\frac{1}{2^{11}} + \frac{1}{3^{11}}2111+31112211+3311\frac{2}{2^{11}} + \frac{3}{3^{11}}2112+3113ExplanationGiven sequence 0,56,536,35216,...0, \frac{5}{6}, \frac{5}{36}, \frac{35}{216}, ...0,65,365,21635,... Each term if expanded becomes u1=0=120−130=121−1+(−1)1⋅13(1−1)u_1 = 0 = \frac{1}{2^0} - \frac{1}{3^0} = \frac{1}{2^{1-1}} + (-1)^1 \cdot \frac{1}{3^{(1-1)}}u1=0=201−301=21−11+(−1)1⋅3(1−1)1u2=56=121+131=122−1+(−1)2⋅13(2−1)u_2 = \frac{5}{6} = \frac{1}{2^1} + \frac{1}{3^1} = \frac{1}{2^{2-1}} + (-1)^2 \cdot \frac{1}{3^{(2-1)}}u2=65=211+311=22−11+(−1)2⋅3(2−1)1u3=536=122−132=123−1+(−1)3⋅13(3−1)u_3 = \frac{5}{36} = \frac{1}{2^2} - \frac{1}{3^2} = \frac{1}{2^{3-1}} + (-1)^3 \cdot \frac{1}{3^{(3-1)}}u3=365=221−321=23−11+(−1)3⋅3(3−1)1u4=35216=123+133=124−1+(−1)4⋅13(4−1)u_4 = \frac{35}{216} = \frac{1}{2^3} + \frac{1}{3^3} = \frac{1}{2^{4-1}} + (-1)^4 \cdot \frac{1}{3^{(4-1)}}u4=21635=231+331=24−11+(−1)4⋅3(4−1)1 Therefore, the nnnth term can be formulated as follows un=12n−1+(−1)n⋅13(n−1)u_n = \frac{1}{2^{n-1}} + (-1)^n \cdot \frac{1}{3^{(n-1)}}un=2n−11+(−1)n⋅3(n−1)1 Then the 121212th term is u12=1212−1+(−1)12⋅13(12−1)u_{12} = \frac{1}{2^{12-1}} + (-1)^{12} \cdot \frac{1}{3^{(12-1)}}u12=212−11+(−1)12⋅3(12−1)1u12=1211+1(13(11))u_{12} = \frac{1}{2^{11}} + 1 \left(\frac{1}{3^{(11)}}\right)u12=2111+1(3(11)1)u12=1211+1311u_{12} = \frac{1}{2^{11}} + \frac{1}{3^{11}}u12=2111+3111
