Question

Find the sum: $1+5+19+65+....$ to $n$ terms.

Answer 1

Solution :-

Sum of $1+5+19+65+...n$ terms

$=(3^1-2^1)+(3^2-2^2)+(3^3-2^3)+...$

$=(3^1+3^2+3^3+...$ up to n terms)-$(2^1+2^2+2^3+...$ up to n terms)

$=\frac{3[3^n-1]}{2}-\frac{2[2^n-1]}{1}$