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Question Number 33152 by Rio Mike last updated on 11/Apr/18

i t i s g i v e n t h a t

\frac{1}{n} \underset{r = 1}{\sum^{n}} x^{r} = 2 a n d \sqrt{\frac{1}{n} \underset{r = 1}{\sum^{n}} {(x_{r})}^{2} - \frac{1}{n^{2}} {(\underset{r = 1}{\sum^{n}})}^{2}} = 3

d e t e r m i n e i n t e r m s o f n t h e v a l u e

o f .

\underset{r = 1}{\sum^{n}} {(x_{r} + 1)}^{2}

Commented by prof Abdo imad last updated on 12/Apr/18

w e h a v e \sum_{r = 1}^{n} x^{r} = 2 n, \frac{1}{n} \sum_{r = 1}^{n} x_{r}^{2} - 1 = 9

b e c a u s e {(Σ 1)}^{2} = n^{2} \Rightarrow \sum_{r = 1}^{n} x_{r}^{2} = 10 n s o

\sum_{r = 1}^{n} {(x_{r} + 1)}^{2} = \sum_{r = 1}^{n} (x_{r}^{2} + 2 x_{r} + 1)

= \sum_{r = 1}^{n} x_{r}^{2} + 2 \sum_{r = 1}^{n} x_{r} + n

= 10 n + 4 n + n = 15 n (i f t h e r e i s n o m i s t a k e i n

t h e q u e s t i o n)