General Term Pattern Recognition

It may be noted that for a mathematical sequence, we can find a generalized formula for the \(n^{th}\) term. There's a theoretical approach, which works in most of the cases. It is based on simple algebra. It states:

If in a sequence, the \(n^{th}\) difference comes out to be a constant, then the general formula for the \(n^{th}\) term is a \(n^{th}\) degree polynomial.

It is basically a converse of the method of finite differences , which states that if we make a sequence of numbers obtained by plugging consecutive numbers in a \(n^{th}\) degree polynomial, then the \(n^{th}\) difference will come out to be a constant.

So, let's understand this with an example.

Find the \( 8^{\text{th}}\) term in the sequence \( 1, 4, 9, 16, 25, ...... \).

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So, we start by taking consecutive differences, and keep on making newer and newer sequences named \(D_1\), \(D_2\) and so on till we get a constant. Let's do it out.

\[S=1, 4, 9, 16, 25, \ldots. \]

Subtracting consecutive terms, \(D_i=S_{i+1}-S_i\) like \(d_1=3\) and so on, we have

\[D_1=3, 5, 7, 9, \ldots.\]

Similarly, we have \[D_2=2, 2, 2, \ldots. \]

YAY!! So \(D_2\) came to be a constant, and so the general \(n^{th}\) term of this sequence is a second degree polynomial. Now we shall find out which one it is.

Let the sequence's general term be defined as \(t_n=An^2+Bn+C\), where \(A, B, C\) are real numbers. We know that \(t_1=1\), \(t_2=4\), and \(t_3=9\). So we have enough equations to solve and find \(A\), \(B\) and \(C\).

We find that \(A=1\), \(B=0\) and \(C=0\).

Kudos! We have generated the \(n^{th}\) term. Now everything is easy. The \(8^{th}\) term is \(8^2\), which is indeed \(64\). \(_\square\)