Question

The Arithmetic Mean of the observations
$\mathrm{1.3.5},\mathrm{3.5.7},\mathrm{5.7.9},...,(2n-1)(2n+1)(2n+3)$ is $(\forall n\in N)$

• A. $2n^3+6n^2+7n-2$
• B. $n^3+8n^2+7n-2$
• C. $2n^3+5n^2+6n-1$
• D. $2n^3+8n^2+7n-2$

Answer 1

The correct option is D $2n^3+8n^2+7n-2$

Given observations $\mathrm{1.3.5},\mathrm{3.5.7},\mathrm{5.7.9},...,(2n-1)(2n+1)(2n+3)$

Mean $=\frac{\sum \left[\right(2n-1\left)\right(2n+1\left)\right(2n+3\left)\right]}{n}$
$=\frac{\sum [8n^3+12n^2-2n-3]}{n}$

Now
$\sum (8n^3+12n^2-2n-3)=8\sum n^3+12\sum n^2-2\sum n-\sum 3$

$=8{\left[\frac{n(n+1)}{2}\right]}^2+12\left[\frac{n(n+1)(2n+1)}{6}\right]-2\left[\frac{n(n+1)}{2}\right]-3n$

$=n\left[2n\right(n+1{)}^2+2(n+1)(2n+1)-(n+1)-3]$

$=n[2n^3+2n+4n^2+4n^2+2+6n-n-4]$
$=n[2n^3+8n^2+7n-2]$
Now mean $=\frac{n[2n^3+8n^2+7n-2]}{n}$
$=2n^3+8n^2+7n-2$