Răspuns:
ex.18
Explicație pas cu pas:
[tex]A = {63}^{n} + {7}^{n + 1}\cdot {3}^{2n + 1} - {21}^{n}\cdot {3}^{n + 2} = \\ = {(7 \cdot {3}^{2} )}^{n} + (7\cdot {7}^{n})\cdot (3\cdot {3}^{2n}) - {(3\cdot 7)}^{n}\cdot ({3}^{2}\cdot {3}^{n}) \\= {7}^{n}\cdot {3}^{2n} + 21\cdot{7}^{n}\cdot {3}^{2n} - {3}^{n}\cdot {7}^{n} \cdot 9\cdot {3}^{n} \\= {7}^{n}\cdot {3}^{2n} + 21\cdot{7}^{n}\cdot {3}^{2n} - 9\cdot {7}^{n} \cdot {3}^{2n} \\ = {7}^{n} \cdot {3}^{2n}\cdot(1 + 21 - 9) = {7}^{n} \cdot {3}^{2n}\cdot 13 \: \: \vdots \ \ \bf \red{13}[/tex]
q.e.d.
(postează separat exercițiile...)
