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 $S=1+2+4+7+\dots .+t_n$
Again, $S=1+2+4+\dots \dots +t_{n-1}+t_n$
On subtracting, we get
$0=1+[1+2+3+\dots \dots (n-1)$ 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,
$S=\frac{n}{2}\left[2t_n+(n-1)1\right]$
$=\frac{n}{2}\left[n^2-n+2+n-1\right]$
$=\frac{n\left(n^2+1\right)}{2}$