Question

$2+12+36+80+...n\mathrm{terms}=?$

• A. $\frac{n\left(n+1\right)\left(n+2\right)\left(3n+5\right)}{24}$
• B. $\frac{n\left(n+1\right)\left(n+2\right)\left(3n+1\right)}{12}$
• C. $\frac{n\left(n+1\right)\left(n+2\right)\left(n+5\right)}{24}$
• D. $\frac{n\left(n+1\right)\left(n+2\right)\left(n+3\right)}{12}$

Answer 1

The correct option is B

$\frac{n\left(n+1\right)\left(n+2\right)\left(3n+1\right)}{12}$

Explanation for correct option:

Step-1: Simplify the given series

Let the given series, $\mathrm{S}=2+12+36+80+...n$

$$\begin{gathered} \Rightarrow \mathrm{S}=\left(1^2+1^3\right)+\left(2^2+2^3\right)+\left(3^2+3^3\right)+......n \\ \Rightarrow \mathrm{S}=\left(1^2+2^2+3^2+.....n\right)+\left(1^3+2^3+3^3+.....n\right) \end{gathered}$$

Step-2: Apply formula for sum of first $n$ natural number and their cubes

$\mathbf{\because }\mathbf{(}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{n}\mathbf{)}\mathbf{=}\frac{\mathbf{n}\mathbf{(}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}\mathbf{(}\mathbf{2}\mathbf{n}\mathbf{+}\mathbf{1}\mathbf{)}}{\mathbf{6}}\mathbf{}\mathbf{and}\mathbf{}\mathbf{(}{\mathbf{1}}^{\mathbf{3}}\mathbf{+}{\mathbf{2}}^{\mathbf{3}}\mathbf{+}{\mathbf{3}}^{\mathbf{3}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{n}\mathbf{)}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{n}\left(n+1\right)}{\mathbf{2}}\mathbf{\right)}}^{\mathbf{2}}$

$$\begin{gathered} \Rightarrow \mathrm{S}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+{\left(\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\right)}^2 \\ \Rightarrow \mathrm{S}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6}+\frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4} \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{\left(2\mathrm{n}+1\right)}{6}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{4}\right) \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{2\left(2\mathrm{n}+1\right)+3\mathrm{n}\left(\mathrm{n}+1\right)}{12}\right) \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{4\mathrm{n}+2+3{\mathrm{n}}^2+3\mathrm{n}}{12}\right) \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{3{\mathrm{n}}^2+7\mathrm{n}+2}{12}\right) \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{3{\mathrm{n}}^2+6\mathrm{n}+\mathrm{n}+2}{12}\right) \\ \Rightarrow \mathrm{S}=\mathrm{n}\left(\mathrm{n}+1\right)\left(\frac{\left(3\mathrm{n}+1\right)\left(\mathrm{n}+2\right)}{12}\right) \\ \Rightarrow \mathrm{S}=\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(\mathrm{n}+2\right)\left(3\mathrm{n}+1\right)}{12} \end{gathered}$$

Therefore, correct answer is option $\left(B\right)$