what is the formula/shortcut for sum of the series where the diff of the consecutive terms is in Geometric progress. like 1+6+31+156+781+... nterms

Hi Jesteen,

The nth term of the above series is $$(1+5+5^2+...+5^{n-1})$$ = $$1*(5^n - 1)/(5-1)$$ = $$(5^n - 1)/4$$

So, the sum till 'n' terms will be (1/4)*$$\sum{5^k}$$ - n/4, where k goes from 1 to n.

When the common ratio of the GP is 'r' instead of 5, the summation is given by $$\frac{1}{r-1}\sum{r^k}$$ - $$\frac{n}{r-1}$$ where k goes from 1 to n