The natural numbers are arranged in rows as f | vedclass

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(B) The number of terms in the $n^{th}$ row is $n$.The last term of the $(n-1)^{th}$ row is the sum of the first $(n-1)$ natural numbers,which is $\fr

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$\frac{n}{2}(n^2 + 1)$

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(B) The number of terms in the $n^{th}$ row is $n$.The last term of the $(n-1)^{th}$ row is the sum of the first $(n-1)$ natural numbers,which is $\frac{(n-1)n}{2}$.Therefore,the first term of the $n^{th}$ row is $\frac{(n-1)n}{2} + 1 = \frac{n^2 - n + 2}{2}$.The terms in the $n^{th}$ row form an arithmetic progression $(A.P.)$ with $n$ terms,first term $a = \frac{n^2 - n + 2}{2}$,and common difference $d = 1$.The sum of the $n$ terms is given by $S_n = \frac{n}{2}[2a + (n-1)d]$.Substituting the values: $S_n = \frac{n}{2}[2(\frac{n^2 - n + 2}{2}) + (n-1)(1)]$.$S_n = \frac{n}{2}[n^2 - n + 2 + n - 1] = \frac{n}{2}(n^2 + 1)$.

The natural numbers are arranged in rows as follows:
$1$
$2, 3$
$4, 5, 6$
$7, 8, 9, 10$
$. . .$
What is the sum of the numbers in the $n^{th}$ row?

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The natural numbers are arranged in rows as follows:$1$$2, 3$$4, 5, 6$$7, 8, 9, 10$$. . .$What is the sum of the numbers in the $n^{th}$ row?