Using the principle of mathematical induction, prove that `(n^(2)+n)` is seven for all ` n in N`.

Using the principle of mathematical induction, prove that `(n^(2)+n)`

1.	Using the principle of mathematical induction, prove that `(n^(2)+n)` is seven for all ` n in N`.
Answer» Let P(n) : `(n^(2)+n)` is even. For n=1, the given expression becomes `(1^(2)+1) = 2`, which is even. So, the given statement is true for n = 1, i.e., P(1) is true. Let P(k) be true. Then, `P(k):(k^(2)+k)` is even ` rArr" " (k^(2)+k) = 2m` for some natural number m. ....(i) Now, ` (k+1)^(2)+(k+1)=k^(2)+3k+2=(k^(2)+k)+2(k+1)` ` = 2m + 2(k+1)" "` [using (i) ] ` =2[m+(k+1)]`, which is clearly even. ` :. P(k+1):(k+1)^(2)+(k+1)` is even. Thus, P(k+1) is true, whenever P(k) is true. `:. ` P(1) is true and P(k+1) is true, whenever P(k) is true. Hence, by the principle of induction, it follows that `(n^(2)+n)` is even for all ` n in N`.