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Question

The $$A.M.$$ of the observations $$1.3.5,3.5,7,9,.,(2n-1)(2n+1) (2n+3)$$ is $$(\forall\ n\ \in\ N)$$


A
2n3+6n2+7n2
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B
n3+8n2+7n2
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C
2n3+5n2+6n2
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D
2n3+8n2+6n3
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Solution

The correct option is D $$2n^{3}+8n^{2}+6n-3$$
The given observation are 1.3.5, 3.5.7.,....,(2n-1)(2n+1)(2n+3).
Arithmetic mean of given Observation are

=$$\sum_{r=1}^n\dfrac{x_i}{n}$$

=$$\sum_{r=1}t_r$$

=$$\sum_{r=1}^n(2n-1)(2n+1)(2n+3)$$

=$$\sum_{r=1}^n(8n^3+12n^2-2n-3)$$

=$$ 8\sum n^3+12\sum n^2-2\sum n-3\sum 1$$ 

=$$8\dfrac{n^2(n+1)^2}{4}+12\dfrac{n}{6}(n+1)(2n+1)-2\dfrac{n}{2}(n+1)-3n$$

=$$n(2(n^3+n+2n^2)+2(2n^2+3n+1)-2n-2-3)$$

=$$n(2n^3+4n^2+2n+4n^2+6n+2-2n-5)$$

=$$n(2n^3+8n^2+6n-3)$$

A.M.=$$\dfrac{n(2n^3+8n^2+6n-3)}{n}$$

=$$2n^3+8n^2+6n-3$$

Mathematics

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