To prove: If nis a positive integer, then n2+2n is either odd or a multiple of 4.
Given Information:
A positive integer n .
Concept Used:
If n is odd then it must be of the form n=2k+1 , where k is a positive integer.
If n is even then it must be of the form n=2k , where k is a positive integer.
Proof:
Let n be a positive integer.
Then n can be either even or odd.
Case 1: Let n be odd.
If n is odd then it must be of the form n=2k+1 , where k is a positive integer.
Substitute n=2k+1 into n2+2n and simplify as follows.
n2+2n=(2k+1)2+2(2k+1)=4k2+1+4k+4k+2=4k2+8k+2+1=2(2k2+4k+1)+1=2m+1 (2k2+4k+1=m)
Thus, if n is odd then n2+2n is also odd.
Case 2: Let n be even.
If n is even then it must be of the form n=2k , where k is a positive integer.
Substitute n=2k into n2+2n and simplify as follows.
n2+2n=(2k)2+2(2k)=4k2+4k=4(k2+k)=4m (k2+k=m)
Thus, if n is even then n2+2n is a multiple of 4.
Therefore, from case 1 and 2 it follows that if nis a positive integer, then n2+2n is either odd or a multiple of 4.
Chapter 1 Solutions
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