Question

The recursive series is defined by the formula ${{t}_{n}}=2{{t}_{n-1}}+3$ and ${{t}_{1}}=-2$, how do you find ${{t}_{6}}$?

Answer 1

From the given question we are asked to find the ${{t}_{6}}$ of the recursive series with given conditions. For the given question we will find the preceding terms using the given conditions. We will find till the term ${{t}_{5}}$ and we will use the substitution method and substitute the ${{t}_{5}}$ in the given formula and the term ${{t}_{6}}$.

Complete step by step solution:
The given recursive series is defined by ${{t}_{n}}=2{{t}_{n-1}}+3$, we are given the first term as ${{t}_{1}}=-2$.
Now we will find the second term form this formula using the given first term. So, we get the following.
$\Rightarrow {{t}_{n}}=2{{t}_{n-1}}+3$
We substitute n as two. So, we get,
$\Rightarrow {{t}_{2}}=2{{t}_{2-1}}+3$
$\Rightarrow {{t}_{2}}=2{{t}_{1}}+3$
$\Rightarrow {{t}_{2}}=2(-2)+3$
$\therefore {{t}_{2}}=-1$
Now we will find the third term form this formula using the second term we got. So, we get the following.
$\Rightarrow {{t}_{n}}=2{{t}_{n-1}}+3$
We substitute n as three. So, we get,
$\Rightarrow {{t}_{3}}=2{{t}_{3-1}}+3$
$\Rightarrow {{t}_{3}}=2{{t}_{2}}+3$
$\Rightarrow {{t}_{3}}=2\left( -1 \right)+3$
$\therefore {{t}_{3}}=1$
Now we will find the fourth term form this formula using the third term we got. So, we get the following.
$\Rightarrow {{t}_{n}}=2{{t}_{n-1}}+3$
We substitute n as four. So, we get,
$\Rightarrow {{t}_{4}}=2{{t}_{4-1}}+3$
$\Rightarrow {{t}_{4}}=2{{t}_{3}}+3$
$\Rightarrow {{t}_{4}}=2+3$
$\therefore {{t}_{4}}=5$
Now we will find the fifth term form this formula using the fourth term we got. So, we get the following.
$\Rightarrow {{t}_{n}}=2{{t}_{n-1}}+3$
We substitute n as five. So, we get,
$\Rightarrow {{t}_{5}}=2{{t}_{5-1}}+3$
$\Rightarrow {{t}_{5}}=2{{t}_{4}}+3$
$\Rightarrow {{t}_{5}}=2\left( 5 \right)+3$
$\therefore {{t}_{5}}=13$
Here we got the fifth term.
Finally the sixth that is the required term will be as follows.
Now we will find the sixth term form this formula using the fifth term we got. So, we get the following.
$\Rightarrow {{t}_{n}}=2{{t}_{n-1}}+3$
We substitute n as six. So, we get,
$\Rightarrow {{t}_{6}}=2{{t}_{6-1}}+3$
$\Rightarrow {{t}_{6}}=2{{t}_{5}}+3$
$\Rightarrow {{t}_{6}}=2\left( 13 \right)+3$
$\therefore {{t}_{6}}=29$

Note: Students must do the calculations correctly. Students should be able to understand series questions accurately. Students should not find the first term and waste time as we are already given it. If we write $\Rightarrow {{t}_{5}}=2{{t}_{5}}+3$ it makes our solution wrong.