$\begingroup$Here, the difference of the digits at level 2 are in G.P., so is there any generalised formulae for calculating nth term if difference of digits at any level are in G.P.
$\endgroup$72 Answers
$\begingroup$The repeated differences are $$ \begin{array}{} 1 & 2 & 5 & 12 & 27 \\ 1 & 3 & 7 & 15 \\ 2 & 4 & 8 \end{array} $$
As you have noticed, the third row is given by $2^n$.
The second row is thus given by the partial sums the third row: $2^n-1$.
The first row is thus given by the partial sums the second row: $2^n-n$.
$\endgroup$1$\begingroup$So, with @Ihf's answer, I think the generalised solution would be that if at any levels difference of terms are in G.P.
$then, n^{th}$ term will be, $$T(n) = \frac{a^{n} + bn + c}{d} $$
where, $a, b, c, d$ are numbers and $n$ is the number of terms.
I, have tested this formula on several cases where difference of terms is in G.P. at any levels, and it is returning me correct answer in all of the cases.
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