# 1952 AHSME Problems/Problem 28

## Problem

In the table shown, the formula relating x and y is: $$\begin{array}{|c|c|c|c|c|c|}\hline x & 1 & 2 & 3 & 4 & 5\\ \hline y & 3 & 7 & 13 & 21 & 31\\ \hline\end{array}$$ $\text{(A) } y = 4x - 1 \qquad\quad \text{(B) } y = x^3 - x^2 + x + 2 \qquad\\ \text{(C) } y = x^2 + x + 1 \qquad \text{(D) } y = (x^2 + x + 1)(x - 1) \qquad\\ \text{(E) } \text{none of these}$

## Solution

A simple method of solving the problem is trying each of the answer choices. One can plug in values of x and y for each, because many x-values and their single corresponding y-values are given.

1. Choice A works for (1,3) and (2,7) but fails to work on (3,13) because 3*4=12 and 12-1=11, not 13.
2. Choice B doesn't work for (2,7) because it would be 8-4+2+2=8, not 7.
3. Choice C actually works for all five pairs, being 1+1+1=3, 4+2+1=7, 9+3+1=13, 16+4+1=21, and 25+5+1=31.

Thus, the answer is $\boxed{\bf{(C)}}$.

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