prove that 7^{2n}+23^{3n-3}*3^{n-1}is divisible by 25 for all natural numbers n... plz say hw to do it..

Let p(n) be the statement given by

**p(n) :** 7^{2n}+23^{3n-3}×3^{n-1} is divisible by 25.

**Step 1:** We have p(1) = 7^{2(1) }+ 23^{3(1)-3 }× 3^{(1)-1}

= 7^{2 }+ 23^{3-3 }× 3^{1-1 }= 49 + 23^{0 }× 3^{0}

= 49 + 1 = 50 which is divisible by 25.

Again, we have p(2) = 7^{2(2) }+ 23^{3(2)-3 }× 3^{(2)-1}

= 7^{4 }+ 23^{6-3 }× 3^{2-1 }= 7^{4 }+ 23^{3 }× 3^{1}

= 2401 + 12167^{ }× 3 = 2401 + 36501 = 38902 which isn't divisible by 25.

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