The odd numbers are divided as follows:Row 1: | vedclass

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(D) The number of terms in the $k^{th}$ row is $2k$. The total number of terms in the first $(n-1)$ rows is $2 + 4 + 6 + \dots + 2(n-1) = 2 	imes \fr

Vedclass · Accepted Answer

$n^3 + (n-1)^3 = 2n^3 - 3n^2 + 3n - 1$

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(D) The number of terms in the $k^{th}$ row is $2k$. The total number of terms in the first $(n-1)$ rows is $2 + 4 + 6 + \dots + 2(n-1) = 2 	imes \frac{(n-1)n}{2} = n(n-1) = n^2 - n$. The first term of the $n^{th}$ row is the $(n^2 - n + 1)^{th}$ odd number. The $m^{th}$ odd number is $2m - 1$. So,the first term $a$ of the $n^{th}$ row is $2(n^2 - n + 1) - 1 = 2n^2 - 2n + 2 - 1 = 2n^2 - 2n + 1$. The $n^{th}$ row is an arithmetic progression with $2n$ terms,first term $a = 2n^2 - 2n + 1$,and common difference $d = 2$. The sum $S_n$ of the $n^{th}$ row is given by $S_n = \frac{2n}{2} [2a + (2n - 1)d]$. $S_n = n [2(2n^2 - 2n + 1) + (2n - 1)2] = n [4n^2 - 4n + 2 + 4n - 2] = n [4n^2] = 4n^3$.

The odd numbers are divided as follows:
Row $1$: $1, 3$
Row $2$: $5, 7, 9, 11$
Row $3$: $13, 15, 17, 19, 21, 23$
Then the sum of the $n^{th}$ row is:

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The odd numbers are divided as follows:Row $1$: $1, 3$Row $2$: $5, 7, 9, 11$Row $3$: $13, 15, 17, 19, 21, 23$Then the sum of the $n^{th}$ row is: