Question

Evaluate the finite sums.

$\sum _{j=11}^{200}{\left(\frac{3}{2}\right)}^j$

Short answer

On evaluating, we get,

$2\left({\left(\frac{3}{2}\right)}^{201}-{\left(\frac{3}{2}\right)}^{11}\right)$

Step-by-step solution

Step 1. Given information.

Consider the given question,

$\sum _{j=11}^{200}{\left(\frac{3}{2}\right)}^j$

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

The geometric series $\sum _{j=11}^{200}{\left(\frac{3}{2}\right)}^j$in the expanded form is $\sum _{j=11}^{200}{\left(\frac{3}{2}\right)}^j={\left(\frac{3}{2}\right)}^{11}+{\left(\frac{3}{2}\right)}^{12}+...+{\left(\frac{3}{2}\right)}^{200}$.

The series has the first term $a={\left(\frac{3}{2}\right)}^{11}$.

The common ratio $r=\frac{3}{2}$with no. of terms,

role="math" localid="1649269759876" $n=190$

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

$$\begin{gathered} S_{190}=\frac{{\left({\displaystyle \frac{3}{2}}\right)}^{11}\left({\left(\frac{3}{2}\right)}^{190}-1\right)}{\left(\frac{2}{3}\right)-1} \\ =\frac{2{\left({\displaystyle \frac{3}{2}}\right)}^{11}\left({\left(\frac{3}{2}\right)}^{190}-1\right)}{3-2} \\ =2\left({\left(\frac{3}{2}\right)}^{201}-{\left(\frac{3}{2}\right)}^{11}\right) \end{gathered}$$