Question

If the natural numbers are written as

Then, the sum of the terms of the $n$th row is

• A. $\frac{n\left(n^{2}-1\right)}{2}$
• B. $\frac{n\left(n^{2}+1\right)}{4}$
• C. $\frac{n\left(n^{2}+1\right)}{2}$
• D. None of these

Answer 1

Answer: (C)

Let $t_{n}$ denotes the $n$th term of the sequence$1,2,4,7$
Let $S=1+2+4+7+\ldots .+t_{n}$
Again, $S=1+2+4+\ldots \ldots+t_{n-1}+t_{n}$
On subtracting, we get
$0=1+[1+2+3+\ldots \ldots(n-1)$ terms $]-t_{n}$
$\Rightarrow t_{n}=1+\frac{n(n-1)}{2}=\frac{n^{2}-n+2}{2}$
which denotes the first term of the $n$th row which contains $n$ terms in A.P., having common difference $1 .$
Hence, we have,
$S =\frac{n}{2}\left[2 t_{n}+(n-1) 1\right]$
$=\frac{n}{2}\left[n^{2}-n+2+n-1\right]$
$=\frac{n\left(n^{2}+1\right)}{2}$