Mathematical Induction (1 Viewer)

[JustRandomThings]

Hey every 1 ty in advance ^^

Prove that 3^(3n) + 2^(n+2) is divisible by 5 for all integers n greater than or equal to 1

 

[pikachu975]

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

 

[leehuan]

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

Not really

Just replace all the parentheses ( ) with braces { }

And use \\ to change lines instead

 

[Drongoski]

Not really

Just replace all the brackets ( ) with braces { }

And use \\ to change lines instead

( ) are parentheses; brackets are [ ].

Of course we often refer to ( ). [ ] and { } simply as brackets.

 

[leehuan]

( ) are parentheses; brackets are [ ].

Of course we often refer to ( ). [ ] and { } simply as brackets.

Fixed. I forgot the word at the time.