Question

Prove that 5n^2 + 3n + 4 is even, for all integers n.

Answer 1

i know that 2n is considered even and 2n + 1 is odd..
i tried to break it down (4n^2 + n^2) + (2n + n) + 4...

Answer 2

Do you know mathematical induction?

Answer 3

learning it, haha

Answer 4

\(\large\color{black}{ \displaystyle 5n^2 + 3n + 4 }\)
\(\large\color{black}{ \displaystyle 5n(n + 3) + 4 }\)
when 'n' is even you get an even 5n times an odd n+3 which is an even number. Then add 4 to this even number -this is still even.
when 'n' is odd, then n+3 is even and 5n is odd, and even n+3 times odd 5n is an even number. Again, add 4 to this even number, and your output is still going to be even.

this is what I would say.

Answer 5

or, instead of analyzing you can break it down as you did,
\(\large\color{black}{ \displaystyle 5n^2 + 3n + 4 }\)
\(\large\color{black}{ \displaystyle 4n^2 + n^2+2n+n + 4 }\)
\(\large\color{black}{ \displaystyle 4n^2 +2n+ n^2+n + 4 }\)
\(\large\color{black}{ \displaystyle 2n(2n +1)+ n(n+1) + 4 }\)

and here, again regardless if you choose an odd or even "n' (as long as it is integer) you get an even output

Answer 6

its beautiful

Answer 7

\(\large\color{black}{ \displaystyle 2n(2n+1)~~~~~~~~+~n(n+1)~~~+4 }\)

\(\large\color{black}{ \displaystyle ({\rm even } \times {\rm odd})~~~~+{\rm even } \times {\rm odd}~~~+{\rm even } }\)

OR, when n is an odd number
\(\large\color{black}{ \displaystyle ({\rm even } \times {\rm odd})~~~~+{\rm odd } \times {\rm even}~~~+{\rm even } }\)

Answer 8

(don't want to do anything complicated on this easy question)

Answer 9

but, you get the point.....

Answer 10

thanks, yeah i just have a hard time developing a "proof-like" answer still, need more practice! Thank you

Answer 11

sure... yw !