Solution to Prove that for any integer n n, if n n is an odd integer, then … - Sikademy
Author Image

Archangel Macsika

Prove that for any integer n n, if n n is an odd integer, then 6n 2 +5n+1 6n2+5n+1 is an even integer.

The Answer to the Question
is below this banner.

Can't find a solution anywhere?

NEED A FAST ANSWER TO ANY QUESTION OR ASSIGNMENT?

Get the Answers Now!

You will get a detailed answer to your question or assignment in the shortest time possible.

Here's the Solution to this Question

Suppose n is odd. Then, n=2k-1, k\in \N

6n^2+5n+1=6(2k-1)^2+5(2k-1)+1\\ =6(4k^2-4k+1)+10k-5+1\\ =24k^2-24k+6+10k-4\\ =24k^2-14k+2\\ =2(12k^2-7k+1)\\ =2M, M=12k^2-7k+1\\ \text{Hence } 6n^2+5n+1 \text{ is even}


Related Answers

Was this answer helpful?

Join our Community to stay in the know

Get updates for similar and other helpful Answers

Question ID: mtid-5-stid-8-sqid-1409-qpid-1147