2. Escreva os seis termos iniciais das sequências dadas pelas seguintes leis a) a_(n)=3n-2,Vngeqslant 1 b) b_(n)=2cdot 3^n,forall ngeqslant 1 C) c_(n)=n(n+1),forall ngeqslant 1 1) d_(n)=(-2)^n,forall ngeqslant 1 e) e_(n)=n^3,Vngeqslant 1

Vamos calcular os seis termos iniciais de cada uma das sequências fornecidas:<br /><br />a) \(a_{n} = 3n - 2, \forall n \geqslant 1\)<br /><br />Para \(n = 1\):<br />\[a_{1} = 3 \cdot 1 - 2 = 1\]<br /><br />Para \(n = 2\):<br />\[a_{2} = 3 \cdot 2 - 2 = 6 - 2 = 4\]<br /><br />Para \(n = 3\):<br />\[a_{3} = 3 \cdot 3 - 2 = 9 - 2 = 7\]<br /><br />Para \(n = 4\):<br />\[a_{4} = 3 \cdot 4 - 2 = 12 - 2 = 10\]<br /><br />Para \(n = 5\):<br />\[a_{5} = 3 \cdot 5 - 2 = 15 - 2 = 13\]<br /><br />Para \(n = 6\):<br />\[a_{6} = 3 \cdot 6 - 2 = 18 - 2 = 16\]<br /><br />Seis termos iniciais: \(1, 4, 7, 10, 13, 16\)<br /><br />b) \(b_{n} = 2 \cdot 3^{n}, \forall n \geqslant 1\)<br /><br />Para \(n = 1\):<br />\[b_{1} = 2 \cdot 3^{1} = 2 \cdot 3 = 6\]<br /><br />Para \(n = 2\):<br />\[b_{2} = 2 \cdot 3^{2} = 2 \cdot 9 = 18\]<br /><br />Para \(n = 3\):<br />\[b_{3} = 2 \cdot 3^{3} = 2 \cdot 27 = 54\]<br /><br />Para \(n = 4\):<br />\[b_{4} = 2 \cdot 3^{4} = 2 \cdot 81 = 162\]<br /><br />Para \(n = 5\):<br />\[b_{5} = 2 \cdot 3^{5} = 2 \cdot 243 = 486\]<br /><br />Para \(n = 6\):<br />\[b_{6} = 2 \cdot 3^{6} = 2 \cdot 729 = 1458\]<br /><br />Seis termos iniciais: \(6, 18, 54, 162, 486, 1458\)<br /><br />c) \(c_{n} = n(n+1), \forall n \geqslant 1\)<br /><br />Para \(n = 1\):<br />\[c_{1} = 1(1+1) = 1 \cdot 2 = 2\]<br /><br />Para \(n = 2\):<br />\[c_{2} = 2(2+1) = 2 \cdot 3 = 6\]<br /><br />Para \(n = 3\):<br />\[c_{3} = 3(3+1) = 3 \cdot 4 = 12\]<br /><br />Para \(n = 4\):<br />\[c_{4} = 4(4+1) = 4 \cdot 5 = 20\]<br /><br />Para \(n = 5\):<br />\[c_{5} = 5(5+1) = 5 \cdot 6 = 30\]<br /><br />Para \(n = 6\):<br />\[c_{6} = 6(6+1) = 6 \cdot 7 = 42\]<br /><br />Seis termos iniciais: \(2, 6, 12, 20, 30, 42\)<br /><br />d) \(d_{n} = (-2)^{n}, \forall n \geqslant 1\)<br /><br />Para \(n = 1\):<br />\[d_{1} = (-2)^{1} = -2\]<br /><br />Para \(n = 2\):<br />\[d_{2} = (-2)^{2} = 4\]<br /><br />Para \(n = 3\):<br />\[d_{3} = (-2)^{3} = -8\]<br /><br />Para \(n = 4\):<br />\[d_{4} = (-2)^{4} = 16\]<br /><br />Para \(n = 5\):<br />\[d_{5} = (-2)^{5} = -32\]<br /><br />Para \(n = 6\):<br />\[d_{6} = (-2)^{6} = 64\]<br /><br />Seis termos iniciais: \(-2, 4, -8, 16, -32, 64\)<br /><br />e) \(e_{n} = n^{3}, \forall n \geqslant 1\)<br /><br />Para \(n = 1