Nth Term

Let the polynomial be:

$y = ax^4 + bx^3 + cx^2 + dx + e$

Now, the function is evaluated at six points and there are five unknown coefficients in the polynomial. So five out of these six points can be used to evaluate the unknown coefficients.

So: $ax_n^4 + bx_n^3 + cx_n^2 + dx_n + e = y_n$ $n \in (1,2,3,4,5)$ $x_n = n$

Writing out the first five equations and rearranging them in a matrix form:

$\left[ \begin{matrix} x_1^4&x_1^3&x_1^2&x_1&1\\x_2^4&x_2^3&x_2^2&x_2&1\\x_3^4&x_3^3&x_3^2&x_3&1\\x_4^4&x_4^3&x_4^2&x_4&1\\x_5^4&x_5^3&x_5^2&x_5&1\end{matrix}\right] \left[\begin{matrix} a\\b\\c\\d\\e \end{matrix}\right] = \left[\begin{matrix} y_1\\y_2\\y_3\\y_4\\y_5 \end{matrix}\right]$ $Ac = b$ $\implies c = A^{-1}b$ $y_1 = 1 \ ; \ y_2 = 1 \ ; \ y_3 = 5 \ ; \ y_4 = 13 \ ; \ y_5 = 41$

The system of linear equations can also be solved by hand easily. After solving, the following two values were computed:

$f(6) = \left[ \begin{matrix} x_6^4&x_6^3&x_6^2&x_6&1 \end{matrix} \right]\left[\begin{matrix} a\\b\\c\\d\\e \end{matrix}\right] =121$ $f(7) = \left[ \begin{matrix} x_7^4&x_7^3&x_7^2&x_7&1 \end{matrix} \right]\left[\begin{matrix} a\\b\\c\\d\\e \end{matrix}\right] =301$

The calculation of $f(6)$ was a check to see if the system of equations were solved correctly and the calculation of $f(7)$ gives the required answer.