Chapter 12.3, Problem 80AYU

To determine

To find: The sum of first fifty number if the sequences is $\left\{{\left(-1\right)}^n\right\}$ .

Answer to Problem 80AYU

The sum of first fifty number if the sequences is $\left\{{\left(-1\right)}^n\right\}$

  $=0$ .

Explanation of Solution

Given information:

The given sequence is $\left\{{\left(-1\right)}^n\right\}$ .

Calculation:

Calculate the sequence,

  $\left\{s_n\right\}=\left\{{\left(-1\right)}^n\right\}$

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

  $\begin{array}{l}\left\{s_n\right\}=\left\{{\left(-1\right)}^n\right\}\\ s_{n-1}={\left(-1\right)}^{\left(n-1\right)}\end{array}$

The difference of the above term is,

  $\begin{array}{l}d=s_n-s_{n-1}\\ \text{ =}\left(-1\right)-{\left(-1\right)}^{n-1}\\ \text{ =}\left(-1\right)+{\left(-1\right)}^n\\ \text{ =2}{\left(-1\right)}^n\end{array}$

The difference of successive term is not a constant, Therefore, the sequence is not arithmetic.

The ratio of the above term is,

  $\begin{array}{l}r=\frac{s_n}{s_{n-1}}\\ \text{ =}\frac{{\left(-1\right)}^n}{{\left(-1\right)}^{n-1}}\\ \text{ =}{\left(-1\right)}^{n-n+1}\\ {\text{ =-1}}^{}\end{array}$

The ratio of the successive term is nonzero constant -1, therefore $\left\{s_n\right\}$ is a geometric sequence with common ratio $-1$ .

Therefore,

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

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

  $\begin{array}{c}{s}_{50}=\frac{\left[{\left(-1\right)}^{50}-1\right]}{2}\\ =\frac{\left[1-1\right]}{2}\\ =0\end{array}$

Therefore, the sum of first fifty number if the sequences is $\left\{{\left(-1\right)}^n\right\}$

  $=0$ .