simpleman wrote:statement: 1+2+...n = (n^2)-n+1

test case: n=1, --> LHS = 1; RHS = 1 --> TRUE.

let n = k+1 -->

LHS = [1+2+...k] + (k+1)

LHS = [(k^2) - k + 1] + (k+1)

LHS = (k^2) + 2 --> simplified.

RHS = ((k+1)^2) - (k+1) + 1

RHS = ((k^2) + 2k + 1) - k

RHS = (k^2) + k + 1 --> simplified.

As can be seen, the simplified LHS does not equal the simplified RHS. Thus, the inductive method cannot prove this statement to be true for all natural numbers. Not that it proves the statement to be false, but since it cannot prove the statement to be true, I take it that I can safely say it is not true for all natural numbers.

The assumption step, for n = k, appears to have been omitted in the above.

After stating the base case (here, the case for n = 1), it is then necessary to make the initial assumption:

For n = k, assume that 1 + 2 + 3 + ... + (n - 1) + n = n^{2} - n + 1.

Only then does one move on to the inductive step. One may use the "n = k" assumption only if one has

*stated* (has explicitly made) that assumption.

In what you have posted, the work is correct but incomplete. You move from "LHS: k

^{2} + 2 and RHS: k

^{2} + k + 1" to "LHS does not equal RHS", but do not justify this move. For completeness, it is necessary that you prove that "2" does not equal "k + 1". Since your base step was for "n = 1", then you need only add that your "n = k" assumption step is "for k a natural number, k > 1". Then, since k > 1, then k

> 2, so k

^{2} > 4, and the conclusion follows.