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Q.

The A.M. of the observations 1.3.5, 3.5.7, 5.7.9, …, (2n – 1) (2n + 1) (2n + 3) is

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a

2n3+6n2+7n−2

b

n3+8n2+7n−2

c

2n3+5n2+6n−1

d

2n3+8n2+7n−2

answer is D.

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Detailed Solution

A.M =1n[1.3.5+3.5.7+…+(2n−1)(2n+1)(2n+3)]                     =1n∑r=1n (2r−1)(2r+1)(2r+3)=∣1n∑r=1n 8r3+12r2−2r−3=1n8n(n+1)22+12n(n+1)(2n+1)6−2n(n+1)2−3n=2n(n+1)2+2(n+1)(2n+1)−(n+1)−3=2n3+8n2+7n−2.
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