Question

Find the missing numbers in the following series:

2,9,28,65_217.

Answer 1

$$\begin{gathered} \mathrm{The}\mathrm{given}\mathrm{series}\mathrm{is}, \\ 2,9,28,65,\boxed{},217 \\ \mathrm{The}\mathrm{numbers}\mathrm{of}\mathrm{the}\mathrm{above}\mathrm{series}\mathrm{are}\mathrm{obtained}\mathrm{by}\mathrm{using}\mathrm{the}\mathrm{formula}{\mathrm{n}}^3+1 \\ \mathrm{First}\mathrm{number}={\left(1\right)}^3+1=1+1=2\left(\mathrm{putting}\mathrm{n}=1\right) \\ \mathrm{second}\mathrm{number}={\left(2\right)}^3+1=8+1=9\left(\mathrm{putting}\mathrm{n}=2\right) \\ \mathrm{third}\mathrm{number}={\left(3\right)}^3+1=27+1=28\left(\mathrm{putting}\mathrm{n}=3\right) \\ \mathrm{fourth}\mathrm{number}={\left(4\right)}^3+1=64+1=65\left(\mathrm{putting}\mathrm{n}=4\right) \\ \mathrm{fifth}\mathrm{number}={\left(5\right)}^3+1=125+1=126\left(\mathrm{putting}\mathrm{n}=5\right) \\ \mathrm{sixth}\mathrm{number}={\left(6\right)}^3+1=216+1=217\left(\mathrm{putting}\mathrm{n}=6\right) \\ \mathrm{Hence},\mathrm{the}\mathrm{missing}\mathrm{number}\mathrm{is}126 \end{gathered}$$