0000:00:00:00:00Start7Number 7ExplanationGiven a sequence 0,34,316,964,...0, \frac{3}{4}, \frac{3}{16}, \frac{9}{64}, ...0,43,163,649,..., then the 121212th term of this sequence is....1211+1212\frac{1}{2^{11}} + \frac{1}{2^{12}}2111+21211211+1222\frac{1}{2^{11}} + \frac{1}{2^{22}}2111+22211211\frac{1}{2^{11}}21111222\frac{1}{2^{22}}22213212\frac{3}{2^{12}}2123ExplanationGiven sequence 0,34,316,964,...0, \frac{3}{4}, \frac{3}{16}, \frac{9}{64}, ...0,43,163,649,... Each term if expanded becomes u1=0=120−120=121−1+(−1)1⋅122(1−1)u_1 = 0 = \frac{1}{2^0} - \frac{1}{2^0} = \frac{1}{2^{1-1}} + (-1)^1 \cdot \frac{1}{2^{2(1-1)}}u1=0=201−201=21−11+(−1)1⋅22(1−1)1u2=34=121+122=122−1+(−1)2⋅122(2−1)u_2 = \frac{3}{4} = \frac{1}{2^1} + \frac{1}{2^2} = \frac{1}{2^{2-1}} + (-1)^2 \cdot \frac{1}{2^{2(2-1)}}u2=43=211+221=22−11+(−1)2⋅22(2−1)1u3=316=122−124=123−1+(−1)3⋅122(3−1)u_3 = \frac{3}{16} = \frac{1}{2^2} - \frac{1}{2^4} = \frac{1}{2^{3-1}} + (-1)^3 \cdot \frac{1}{2^{2(3-1)}}u3=163=221−241=23−11+(−1)3⋅22(3−1)1u4=964=123+126=124−1+(−1)4⋅122(4−1)u_4 = \frac{9}{64} = \frac{1}{2^3} + \frac{1}{2^6} = \frac{1}{2^{4-1}} + (-1)^4 \cdot \frac{1}{2^{2(4-1)}}u4=649=231+261=24−11+(−1)4⋅22(4−1)1 Therefore, the nnnth term can be formulated as follows un=12n−1+(−1)n⋅122(n−1)u_n = \frac{1}{2^{n-1}} + (-1)^n \cdot \frac{1}{2^{2(n-1)}}un=2n−11+(−1)n⋅22(n−1)1 Then the 121212th term is u12=1212−1+(−1)12⋅122(12−1)u_{12} = \frac{1}{2^{12-1}} + (-1)^{12} \cdot \frac{1}{2^{2(12-1)}}u12=212−11+(−1)12⋅22(12−1)1u12=1211+1(122(11))u_{12} = \frac{1}{2^{11}} + 1 \left(\frac{1}{2^{2(11)}}\right)u12=2111+1(22(11)1)u12=1211+1222u_{12} = \frac{1}{2^{11}} + \frac{1}{2^{22}}u12=2111+2221
7Number 7ExplanationGiven a sequence 0,34,316,964,...0, \frac{3}{4}, \frac{3}{16}, \frac{9}{64}, ...0,43,163,649,..., then the 121212th term of this sequence is....1211+1212\frac{1}{2^{11}} + \frac{1}{2^{12}}2111+21211211+1222\frac{1}{2^{11}} + \frac{1}{2^{22}}2111+22211211\frac{1}{2^{11}}21111222\frac{1}{2^{22}}22213212\frac{3}{2^{12}}2123ExplanationGiven sequence 0,34,316,964,...0, \frac{3}{4}, \frac{3}{16}, \frac{9}{64}, ...0,43,163,649,... Each term if expanded becomes u1=0=120−120=121−1+(−1)1⋅122(1−1)u_1 = 0 = \frac{1}{2^0} - \frac{1}{2^0} = \frac{1}{2^{1-1}} + (-1)^1 \cdot \frac{1}{2^{2(1-1)}}u1=0=201−201=21−11+(−1)1⋅22(1−1)1u2=34=121+122=122−1+(−1)2⋅122(2−1)u_2 = \frac{3}{4} = \frac{1}{2^1} + \frac{1}{2^2} = \frac{1}{2^{2-1}} + (-1)^2 \cdot \frac{1}{2^{2(2-1)}}u2=43=211+221=22−11+(−1)2⋅22(2−1)1u3=316=122−124=123−1+(−1)3⋅122(3−1)u_3 = \frac{3}{16} = \frac{1}{2^2} - \frac{1}{2^4} = \frac{1}{2^{3-1}} + (-1)^3 \cdot \frac{1}{2^{2(3-1)}}u3=163=221−241=23−11+(−1)3⋅22(3−1)1u4=964=123+126=124−1+(−1)4⋅122(4−1)u_4 = \frac{9}{64} = \frac{1}{2^3} + \frac{1}{2^6} = \frac{1}{2^{4-1}} + (-1)^4 \cdot \frac{1}{2^{2(4-1)}}u4=649=231+261=24−11+(−1)4⋅22(4−1)1 Therefore, the nnnth term can be formulated as follows un=12n−1+(−1)n⋅122(n−1)u_n = \frac{1}{2^{n-1}} + (-1)^n \cdot \frac{1}{2^{2(n-1)}}un=2n−11+(−1)n⋅22(n−1)1 Then the 121212th term is u12=1212−1+(−1)12⋅122(12−1)u_{12} = \frac{1}{2^{12-1}} + (-1)^{12} \cdot \frac{1}{2^{2(12-1)}}u12=212−11+(−1)12⋅22(12−1)1u12=1211+1(122(11))u_{12} = \frac{1}{2^{11}} + 1 \left(\frac{1}{2^{2(11)}}\right)u12=2111+1(22(11)1)u12=1211+1222u_{12} = \frac{1}{2^{11}} + \frac{1}{2^{22}}u12=2111+2221
