find the sum to n series 3x1^2+5x2^2+7x3^2+..........

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$$\begin{gathered} \mathrm{S}=3.1^2+5.2^2+7.3^2+...\mathrm{upto} \text{'}\mathrm{n}\text{'} \mathrm{terms} \\ 3,5,7 \mathrm{is} \mathrm{an} \mathrm{A}.\mathrm{P}. \mathrm{of} \mathrm{first} \mathrm{term}\left(\mathrm{a}\right)=3 \mathrm{and} \mathrm{common} \mathrm{difference}\left(\mathrm{d}\right)=5-3=2 \\ \mathrm{We} \mathrm{know} \mathrm{that}: \\ {\mathrm{T}}_{\mathrm{r}}=\mathrm{a}+\left(\mathrm{r}-1\right)\mathrm{d} \\ =3+\left(\mathrm{r}-1\right)2 \\ =3+2\mathrm{r}-2 \\ =2\mathrm{r}+1 \\ \mathrm{Hence} \mathrm{given} \mathrm{series} \mathrm{can} \mathrm{be} \mathrm{written} \mathrm{as}: \\ \mathrm{S}=\sum _{\mathrm{r}=1}^{\mathrm{n}} \left(2\mathrm{r}+1\right) {\mathrm{r}}^2 \\ =\sum _{\mathrm{r}=1}^{\mathrm{n}} 2{\mathrm{r}}^3+\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{r}}^2 \\ =2\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{r}}^3+\sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{r}}^2 \\ \mathrm{Writing} \mathrm{respective} \mathrm{values} \mathrm{of} \sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{r}}^3 \mathrm{and} \sum _{\mathrm{r}=1}^{\mathrm{n}} {\mathrm{r}}^2 \mathrm{we} \mathrm{get}: \\ \mathrm{S}=2 \frac{{\mathrm{n}}^2{\left(\mathrm{n}+1\right)}^2}{4}+\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(2\mathrm{n}+1\right)}{6} \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left({\mathrm{n}}^2+\mathrm{n}+\frac{2\mathrm{n}+1}{3}\right) \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)}{2}\left(\frac{3{\mathrm{n}}^2+3\mathrm{n}+2\mathrm{n}+1}{3}\right) \\ =\frac{\mathrm{n}\left(\mathrm{n}+1\right)\left(3{\mathrm{n}}^2+5\mathrm{n}+1\right)}{6} \\ \mathbf{\Rightarrow }\mathbf{3}\mathbf{.}{\mathbf{1}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{.}{\mathbf{2}}^{\mathbf{2}}\mathbf{+}\mathbf{7}\mathbf{.}{\mathbf{3}}^{\mathbf{2}}\mathbf{+}\mathbf{.}\mathbf{.}\mathbf{.}\mathbf{upto}\mathbf{ }\mathbf{\text{'}}\mathbf{n}\mathbf{\text{'}}\mathbf{ }\mathbf{terms}\mathbf{=}\frac{\mathbf{n}\left(\mathbf{n}\mathbf{+}\mathbf{1}\right)\left(\mathbf{3}{\mathbf{n}}^{\mathbf{2}}\mathbf{+}\mathbf{5}\mathbf{n}\mathbf{+}\mathbf{1}\right)}{\mathbf{6}} \end{gathered}$$

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