This answer is restricted. Please login to view the answer of this question.Login Now
Let p(n) = n2 + n
First we perform basic steps
We show p(1) is true
i.e. p(1) = 12 + 1 = 2 (which is divisible by 2)
Hence, n2 + n is an even integer when n = 1
Let us suppose, for any arbitary k, p(k) is true. i,e, it is divisible by 2 so that it is an even integer.
i.e. k2 + k = 2t, where t is an integer
Now, we have to show p(k + 1) is also even i.e. divisible by 2
(k + 1)2 + (k + 1)
= k2 + 2k + 1 + k + 1
= k2 + k + 2k + 2
= 2t + 2k + 2
= 2t + 2(k+1)
= 2t + 2m (where m is an integer)
= 2s (where is an integer)
Here, 2s is divisible by 2. Som p(x + 1) is an even integer. So, its true.
Hence, by matematical induction, n2 + n is even whenever p ≥ 1
Click here to submit your answer.