Answer:
[tex]{ \rm{ \frac{ {4}^{a + 2} + {4}^{a} }{17 \times {4}^{a} } }} \\ \\ = { \rm{ \frac{( {4}^{a} \times {4}^{2} ) + {4}^{a} }{17 \times {4}^{a} } }} \\ \\ = { \rm{ \frac{ {4}^{a} }{ {4}^{a} } ( \frac{ {4}^{2} + 1}{17}) }} \\ \\ = { \rm{ \frac{ {4}^{2} + 1 }{17} }} \\ \\ = { \rm{ \frac{16 + 1}{17} }} \\ \\ { \rm{ = \frac{17}{17} }} \\ \\ = 1[/tex]
hence proved
