Question

Write the first five terms sequence defined by ${{a}_{n}}=n\dfrac{{{n}^{2}}+5}{4}$ .

Answer 1

Hint: To find the first five terms, we have to substitute different values of n in the sequence ${{a}_{n}}=n\dfrac{{{n}^{2}}+5}{4}$ starting from 1 to 5, that is, $n=1,2,3,4,5$ .We have to simplify the result to find the values of ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}$ and ${{a}_{5}}$ .

Complete step by step answer:
We have to find the first five terms, that is ${{a}_{1}},{{a}_{2}},{{a}_{3}},{{a}_{4}}$ and ${{a}_{5}}$ . For this, we have to substitute $n=1,2,3,4,5$ in the sequence ${{a}_{n}}=n\dfrac{{{n}^{2}}+5}{4}$ .
Let us find the first term. We have to substitute $n=1$ in the given sequence.
$\Rightarrow {{a}_{1}}=1\times \dfrac{{{1}^{2}}+5}{4}$
We have to simplify the above expression.
$\Rightarrow {{a}_{1}}=\dfrac{6}{4}$
Let us cancel the common factor 2 from the numerator and the denominator.
$\Rightarrow {{a}_{1}}=\dfrac{{{\require{cancel}\cancel{6}}^{3}}}{{{\require{cancel}\cancel{4}}^{2}}}$
We can write the result of the simplification shown above as
$\Rightarrow {{a}_{1}}=\dfrac{3}{2}$
Now, we have to substitute $n=2$ in the given sequence to get the second term.
$\begin{align}
  & \Rightarrow {{a}_{2}}=2\times \dfrac{{{2}^{2}}+5}{4} \\
 & \Rightarrow {{a}_{2}}=2\times \dfrac{4+5}{4} \\
 & \Rightarrow {{a}_{2}}=\require{cancel}\cancel{2}\times \dfrac{9}{{{\require{cancel}\cancel{4}}^{2}}} \\
 & \Rightarrow {{a}_{2}}=\dfrac{9}{2} \\
\end{align}$
Let us find the third term by substituting $n=3$ in the given sequence.
\[\begin{align}
  & \Rightarrow {{a}_{3}}=3\times \dfrac{{{3}^{2}}+5}{4} \\
 & \Rightarrow {{a}_{3}}=3\times \dfrac{9+5}{4} \\
 & \Rightarrow {{a}_{3}}=3\times \dfrac{{{\require{cancel}\cancel{14}}^{7}}}{{{\require{cancel}\cancel{4}}^{2}}} \\
 & \Rightarrow {{a}_{3}}=\dfrac{21}{2} \\
\end{align}\]
Now, we have to substitute $n=4$ in the given sequence to get the fourth term.
\[\begin{align}
  & \Rightarrow {{a}_{4}}=4\times \dfrac{{{4}^{2}}+5}{4} \\
 & \Rightarrow {{a}_{4}}=\require{cancel}\cancel{4}\times \dfrac{16+5}{\require{cancel}\cancel{4}} \\
 & \Rightarrow {{a}_{4}}=21 \\
\end{align}\]
Finally, to find the fifth term, we have to substitute $n=5$ in the given sequence.
\[\begin{align}
  & \Rightarrow {{a}_{5}}=5\times \dfrac{{{5}^{2}}+5}{4} \\
 & \Rightarrow {{a}_{5}}=5\times \dfrac{25+5}{4} \\
 & \Rightarrow {{a}_{5}}=\dfrac{{{\require{cancel}\cancel{150}}^{75}}}{{{\require{cancel}\cancel{4}}^{2}}} \\
 & \Rightarrow {{a}_{5}}=\dfrac{75}{2} \\
\end{align}\]
Hence, the first five terms of the sequence defined by ${{a}_{n}}=n\dfrac{{{n}^{2}}+5}{4}$ are $\dfrac{3}{2},\dfrac{9}{2},\dfrac{21}{2},21,\dfrac{75}{2}$ .

Note: Students must note that we have not considered the value $n=0$ . This is because we have to find the terms and there is no such term as the zeroth term. Students must simplify the answers as much as possible and cancel the common terms and factors, if any.