Question

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

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

Answer 1

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

=${\sum }_{r=1}^n\frac{x_i}{n}$

=${\sum }_{r=1}{t}_r$

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

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

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

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

=$n\left(2\right(n^3+n+2n^2)+2(2n^2+3n+1)-2n-2-3)$

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

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

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

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