Answer:
\[{{\left( 25 \right)}^{n-1}}+100={{5}^{\left( 2n-1 \right)}}\] \[\therefore \] \[{{\left( {{5}^{2}} \right)}^{n-1}}+100={{5}^{\left( 2n-1 \right)}}\] \[{{\left( {{5}^{2}} \right)}^{2n-2}}+100={{5}^{\left( 2n-1 \right)}}\] \[{{\left( {{5}^{2}} \right)}^{2n-2}}-{{5}^{\left( 2n-1 \right)}}=-100\] \[{{5}^{2}}^{n-1}\times \left( 5-1 \right)=100\] \[{{5}^{2}}^{n-1}\times 4=100\] \[{{5}^{2n-2}}=\frac{100}{4}=25\] Thus, \[{{5}^{2}}^{n-1}={{5}^{2}}\]
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