Use mathematical induction to prove that the statement is true for every positive integer n. Show your work.? 2 is a factor of n^2 - n + 2

2 Answers
Narad T.
May 2, 2018

Please see the explanation below.

Explanation:

Let #f(k)=k^2-k+2#

Let the statement be #P(n), AA n in NN#

Then

#P(1), {f(1)=-1-1+2=2}# and #2# is divisible by #2#

The statement is true for #n=1#

Let the statement be true for #k=n#

#f(n)=n^2-n+2=2p, AA p in NN#

#n^2=2p+n-2#

#f(n)# is divisible by #2#

Then,

#f(n+1)=(n+1)^2-(n+1)+2#

#=n^2+2n+1-n-1+2#

#=n^2+n+2#

#=2p+n-2+n+2#

#=2p+2n#

#=2(p+n)#

Therefore,

#f(n+1)# is divisible by #2#

The statement is true for #P(n+1)#

Conclusion : As the statement is true for #P(1)#, #P(n)# and #P(n+1)#, the statement is true for all values of #n#. #(n^2-n+2)# is divisible by #2# for #AA n in NN#

Shwetank Mauria
May 2, 2018

Please see below.

Explanation:

Induction method is used to prove a statement. Most commonly, it is used to prove a statement, involving, say #n# where #n# represents the set of all natural numbers.

Induction method involves two steps, One, that the statement is true for #n=1# and say #n=2#. Two, we assume that it is true for #n=k# and prove that if it is true for #n=k#, then it is also true for #n=k+1#.

First Step For #n=1#, #n^2-n+2# becomes #1^2-1+2=1-1+2=2# and #2# is a factor of #n^2-n+2# when #n=1#. Also for #n=2#, we have #2^2-2+2=4# and #2# is a factor of #n^2-n+2#.

Second Step Let #2# be factor of #n^2-n+2# for #n=k# i.e. #k^2-k+2=2m#, where #m# is a natural number.

Then for #n=k+1#, we have

#n^2-n+2=(k+1)^2-(k+1)+2#

= #k^2+2k+1-k-1+2#

= #k^2+k+2#

= #k^2-k+2+2k#

= #2m+2k#

= #2(m+k)#

and #2# is a factor of #n^2-n+2# if #n=k+1# too.

Hence, #2# is a factor of #n^2-n+2# for every integer #n#.

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