Question

A sequence of odd positive integer written as $1,3,5,7,9,11,13,15,17,19,21,23,25,27$ the mean of the $n^{th}$ row is

• A. $\frac{n^3(2n^2+1)}{3}$
• B. $\frac{n^3(4n^2+2)}{6}$
• C. $\frac{n(n-1)(2n-1)}{6}$
• D. $\frac{n(2n^2+1)}{3}$

Answer 1

The correct option is D $\frac{n(2n^2+1)}{3}$

The number of numbers in the $n^{th}$ row $=n^2$
Sequence of the first terms in different row is $1,3,11,29,61,..$.
$\therefore$ $T_n$ of 1, 3, 11, 29, 61,...
$=\frac{1}{2}(2n^3-3n^2+n+3)$ $=$ first element of $n^{th}$ row.
Similarly , sequence of last terms of each row $=$ $1,9,27,59,...$
$\therefore$ $t_n=\frac{1}{3}\left[2n^3+3n^2+n-3\right]$ $=$ last element of the $n^{th}$ row
Hence, in the $n^{th}$ row element can be written as
$\frac{1}{3}(2n^3-3n^2+n+3)...\frac{1}{3}(2n^2+3n^2+n-3)$
(Note adding 2 in the preceding number to get the succeeding number).
$\therefore$ Sum of the elements of $n^{th}$ row (using sum of $n$ terms of A .P.)
$=\frac{N}{2}(A+L)=\frac{n^2}{2}\left[\frac{4n^2+2n}{3}\right]=\frac{n^3}{3}(2n^2+1)$
$\therefore$ mean of the number in the $n^{th}$ row
$=n^3\frac{(2n^2+1)}{3\times n^2}=\frac{n(2n^2+1)}{3}$