Question

Find first five terms of the sequence in which $a_1=3,a_{n+1}=2a_n+5$.

Short answer

The first five terms of the sequence are$3,11,27,59\text{and}123$.

Step-by-step solution

Step 1. Write down the given information.

The given sequence has $a_1=3 \text{and}{a}_{n+1}=2a_n+5$.

Step 2. Evaluating first five terms of the given sequence.

Since, the first term is $a_1=3$ of the given sequence. Therefore, the remaining four terms can be obtained by plugging $n=1,2,3,4$ to get $a_2,a_3,a_4\text{and}{a}_5$ in the given sequence $a_{n+1}=2a_n+5$.

For $n=1$,

$\begin{array}{c}{a}_{n+1}=2a_n+\mathrm{5....}\left(\text{Given}\right)\\ a_{\left(1\right)+1}=2a_{\left(1\right)}+\mathrm{5....}\left(\text{For}n=1\right)\\ a_2=2\left(3\right)+\mathrm{5....}\left(\because a_1=3\right)\\ a_2=11\end{array}$

For $n=2$,

$\begin{array}{c}{a}_{n+1}=2a_n+\mathrm{5....}\left(\text{Given}\right)\\ a_{\left(2\right)+1}=2a_{\left(2\right)}+\mathrm{5....}\left(\text{For}n=2\right)\\ a_3=2\left(11\right)+\mathrm{5....}\left(\because a_2=11\right)\\ a_3=27\end{array}$

For $n=3$,

$\begin{array}{c}{a}_{n+1}=2a_n+\mathrm{5....}\left(\text{Given}\right)\\ a_{\left(3\right)+1}=2a_{\left(3\right)}+\mathrm{5....}\left(\text{For}n=3\right)\\ a_4=2\left(27\right)+\mathrm{5....}\left(\because a_3=27\right)\\ a_4=59\end{array}$

For $n=4$,

$\begin{array}{c}{a}_{n+1}=2a_n+\mathrm{5....}\left(\text{Given}\right)\\ a_{\left(4\right)+1}=2a_{\left(4\right)}+\mathrm{5....}\left(\text{For}n=4\right)\\ a_5=2\left(59\right)+\mathrm{5....}\left(\because a_4=59\right)\\ a_5=123\end{array}$

Since, $a_1=3,a_2=11,a_3=27,a_4=\text{59 and}{a}_5=123$. Therefore, the first five terms of the sequence are $3,11,27,59\text{and}123$.

Step 3. Conclusion.

The first five terms of the sequence are $3,11,27,59\text{and}123$.