Question

Find the sum of the first n terms of the special series :
1 + (1+3)+(1+3+3^2)+(1+3+3^2+3^3)+ -------- upto n terms

Answer 1

\[ \large ...=\sum_{i=1}^n\left(\sum_{j=0}^{i-1}3^j\right) \]

Answer 2

here an=(1+3+3^2+3^3+.....3^(n-1))
so an=(3^n -1)/2 =.5 *3^n -.5
hence S= .5 (3+3^2+3^3+.....3^(n)) -.5(1+1+1+...ntimes)
=.5 *3*(3^n -1)/(3-1) -.5n =.75 ((3^n -1) -.5n

Answer 3

\[ \large =n+(n-1)3+(n-2)3^2+\dots+3^n \]

Answer 4

sry it is 3^2