3^(3n) + 2^(n+2)

Step 1: Prove true for n=1

LHS = 3^(3) + 2^(1+2)

= 27 + 8

= 35 which is divisible by 5, therefore true for n=1

Step 2: Assume true for n=k

3^(3k) + 2^(k+2) = 5M (where M is an integer)

Step 3: Prove true for n=k+1

LHS = 3^(3(k+1)) + 2^(k+3)

= 3^(3k+3) + 2^(k+2+1)

= 3^3 * 3^(3k) + 2 * 2^(k+2)

= 27 (3^(3k) + 2^(k+2)) - 25(2^(k+2)) ----- I factorised 27(2^(k+2)) so I had to minus 25 of it to pay it back

= 27 (5M) - 25(2^(k+2)) ----- using assumption

= 5 (27M - 5(2^(k+2))) and since 27, M, 5, and 2^(k+2) are integers,

then it is true for n=k+1

Step 4: Conclusion

By mathematical induction, 3^(3n) + 2^(n+2) is divisible by 5 for n >= 1

Soz if it's hard to read it'd take ages to type this in latex

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