Chapter 12.3, Problem 79AYU

To determine

To find: The sum of first fifty number if the sequences is $\left\{3^{n/2}\right\}$ .

Answer to Problem 79AYU

The sum of first fifty number if the sequence is $\left\{3^{n/2}\right\}$

  $=2.004706374\times {10}^{12}$ .

Explanation of Solution

Given information:

The given sequence is $\left\{3^{n/2}\right\}$ .

Calculation:

Calculate the sequence,

  $\left\{e_n\right\}=\left\{3^{\frac{3}{2}}\right\}$

The $n^{th}$ term and the ${\left(n-1\right)}^{th}$ term are

  $\begin{array}{l}\left\{e_n\right\}=\left\{3^{\frac{3}{2}}\right\}\\ e_{n-1}=3^{\frac{n-1}{2}}\end{array}$

The ratio of the above term is,

  $\begin{array}{l}r=\frac{e_n}{e_{n-1}}\\ \text{ =}\frac{3^{\frac{n}{2}}}{3^{\frac{n-1}{2}}}\\ {\text{ =3}}^{\frac{n-n+1}{2}}\\ {\text{ =3}}^{\frac{1}{2}}\end{array}$

The ratio of successive term is nonzero constant, Therefore, the sequence $\left\{3^{n/2}\right\}$ is a geometric sequence with common ratio $3^{\frac{1}{2}}$ .

  $\left|r\right|>1$ ,

Therefore,

  $\begin{array}{c}{s}_n=\frac{a_1\left(r^n-1\right)}{\left(r-1\right)}\\ =\frac{3^{\frac{1}{2}}\left[\left(3^{\frac{1}{2}}\right)-1\right]}{\left(3^{\frac{1}{2}}-1\right)}\\ \end{array}$

The sum of first fifty terms hence, $n=50$

  $\begin{array}{c}{s}_{50}=\frac{\sqrt{3}\left[{\left(\sqrt{3}\right)}^{50}-1\right]}{\left(\sqrt{3}-1\right)}\\ =2.004706374\end{array}$

Therefore, The sum of first fifty number if the sequence is $\left\{3^{n/2}\right\}$

  $=2.004706374\times {10}^{12}$ .