Question

The arithmetic mean (A.M.) of the observations $\mathrm{1.3.5},\mathrm{3.5.7},\mathrm{5.7.9}...(2n–1).(2n+1).(2n+3)$ is

• 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

$2n^3+8n^2+7n–2$

Use $\overline{x}=\frac{x_1+x_2+...+x_n}{n}=\frac{1}{n}{\sum }_{r=1}^n{x}_r$

A.M. $=\frac{1}{n}[\mathrm{1.3.5}+\mathrm{3.5.7}+....+(2n-1\left)\right(2n+1\left)\right(2n+3\left)\right]$

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

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

$=2n(n+1{)}^2+2(n+1\left)\right(2n+1)-(n+1)-3=2n^3+8n^2+7n-2$