Solution

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.

We have \(u_n = a+bn+c2^n,\) so \[\begin{align} a+b+2c&=2, \label{eq:1}\\ a+2b+4c&=-1, \label{eq:2}\\ a+3b+8c&=3. \label{eq:3} \end{align}\]

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.

This will be \[\begin{align*} (a + b + 2c) + (a + 2b + 4c) + &(a + 3b + 8c) +...+ (a+nb+2^nc)\\ &= na + b(1 + 2 + .. + n) + c( 2 + 2^2 + 2^3 + ... + 2^n)\\ &= na + b\dfrac{n(n+1)}{2}+ 2c(2^{n}-1)\\ &= 5n -5n^2 - 5n +7\times 2^n -7\\ &= -5n^2 +7\times 2^n -7. \end{align*}\]

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