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If the sum of squares of the deviations from the mean of the data $x_i,(i=1,2, \ldots, n)$ is $n^2$, where $\bar{x}$ is the mean of $x_1$ 's, then the sum of squares of $x_i$ 's is
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$2 n \mathrm{x}^{-2}$
Given, Sum of squares of deviations from the mean of data $x_i=n \bar{x}^2$
$\begin{aligned} & \Rightarrow \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2+n \bar{x}^2-2 \bar{x} \sum_{i=1}^n x_i=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2-2 \bar{x} \cdot n \bar{x}=0 \quad\left\{\because \bar{x}=\frac{\sum x_i}{n}\right\} \\ & \Rightarrow \sum_{i=1}^n x_i^2=2 n \bar{x}^2\end{aligned}$
$\begin{aligned} & \Rightarrow \sum_{i=1}^n\left(x_i-\bar{x}\right)^2=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2+n \bar{x}^2-2 \bar{x} \sum_{i=1}^n x_i=n \bar{x}^2 \\ & \Rightarrow \sum_{i=1}^n x_i^2-2 \bar{x} \cdot n \bar{x}=0 \quad\left\{\because \bar{x}=\frac{\sum x_i}{n}\right\} \\ & \Rightarrow \sum_{i=1}^n x_i^2=2 n \bar{x}^2\end{aligned}$
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