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(B) The $n^{th}$ row contains $n$ terms. 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}

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

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(B) The $n^{th}$ row contains $n$ terms. 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 $\frac{n(n-1)}{2} + 1 = \frac{n^2 - n + 2}{2}$.The $n^{th}$ row is 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 $n$ terms of an $A.P.$ 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 written as follows. The sum of the numbers in the $n^{th}$ row is:
$1$
$2$ $3$
$4$ $5$ $6$
$7$ $8$ $9$ $10$
$. . .$

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