Question

Evaluate the finite sums.

$\sum _{j=1}^{60}{3}^j$

Short answer

On evaluating, we get,

$\frac{3\left(3^{60}-1\right)}{3-1}$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{j=1}^{60}{3}^j$

Step 2. Use the sum of n terms of geometric series.

The geometric series $\sum _{j=1}^{60}{3}^j$in the expanded form is $\sum _{j=1}^{60}{3}^j=3+3^2+3^3+...+3^{60}$.

The series has the first term $a=1$.

The common ratio $r=3$with no. of terms,

$n=60$

The finite sum of the geometric series with ratio greater than $1$, localid="1649266007528" $S_n=\frac{a\left(r^n-1\right)}{r-1}$.

Substituting the values,

$S_{60}=\frac{3\left(3^{60}-1\right)}{3-1}$