The \(n\)th term of a certain series is of the form \(a+bn+c2^n\), where \(a\), \(b\), and \(c\) are numbers. If the first three terms are \(2\), \(-1\), and \(3\), find the values of \(a\), \(b\), and \(c\), and the sum of the first \(n\) terms.

Taking \(\eqref{eq:2}\)\(-\)\(\eqref{eq:1}\) gives \(b+2c=-3\), while \(\eqref{eq:3}\)\(-\)\(\eqref{eq:2}\) gives \(b + 4c = 4\).

Subtracting these two equations gives \(2c = 7\), and so \(c = \dfrac{7}{2}\).

Substituting back into the earlier equations gives \(b = -10\), and \(a = 5\).

So the \(n^{th}\) term \(u_n = 5-10n+\dfrac{7}{2}2^n= 5-10n+7\times2^{n-1}.\)

Check for the third term: does \(3 = 5 - 30 + 28\)? Yes.

… and the sum of the first \(n\) terms.

Check: when \(n= 3\), the sum of the first \(n\) terms is \(4\). Our formula gives \(-45+56-7 = 4\), so we have agreement.