Please help me with 12th sum
correct option is D

Dear Student,
Please find below the solution to the asked query:

$$\begin{gathered} \sum _{\mathrm{r}=0}^{\mathrm{n}} \frac{\mathrm{n}-2\mathrm{r}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}} \\ \mathrm{If} \mathrm{you} \mathrm{expand} \mathrm{it} \mathrm{and} \mathrm{name} \mathrm{terms} \mathrm{as} {\mathrm{T}}_0,{\mathrm{T}}_1,...{\mathrm{T}}_{\mathrm{n}}, \mathrm{then} \\ {\mathrm{T}}_0=\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_0}=\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \left[\mathrm{As}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}\right] \\ {\mathrm{T}}_{\mathrm{n}}=\frac{\mathrm{n}-2\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}}=-\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}} \\ \Rightarrow {\mathrm{T}}_0+{\mathrm{T}}_{\mathrm{n}}=\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}}-\frac{\mathrm{n}}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}}}=0 \\ {\mathrm{T}}_1=\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_1}=\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}} \left[\mathrm{As}{ }^{\mathrm{n}}{\mathrm{C}}_{\mathrm{r}}{=}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-\mathrm{r}}\right] \\ {\mathrm{T}}_{\mathrm{n}-1}=\frac{\mathrm{n}-2\mathrm{n}+2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}=\frac{-\mathrm{n}+2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}=-\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}} \\ \Rightarrow {\mathrm{T}}_1+{\mathrm{T}}_{\mathrm{n}-1}=\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}-\frac{\mathrm{n}-2}{{}^{\mathrm{n}}{\mathrm{C}}_{\mathrm{n}-1}}=0 \\ \mathrm{Similarly} \mathrm{you} \mathrm{will} \mathrm{see} \mathrm{that} \mathrm{each} \mathrm{pair} \mathrm{will} \mathrm{cancel} \mathrm{out} \mathrm{each} \mathrm{other}, \mathrm{hecne} \\ \mathbf{\sum }_{\mathbf{r}\mathbf{=}\mathbf{0}}^{\mathbf{n}}\mathbf{ }\frac{\mathbf{n}\mathbf{-}\mathbf{2}\mathbf{r}}{{}^{\mathbf{n}}{\mathbf{C}}_{\mathbf{r}}}\mathbf{=}\mathbf{0}\mathbf{ }\mathbf{\to }\mathbf{Option}\left(\mathbf{D}\right) \end{gathered}$$

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