Question

The odd positive numbers are written in the form of a triangle
Find the sum of terms in $n^{th}$ row.

• A. $(n-1{)}^3$
• B. $n^3$
• C. $n^3-1$
• D. $(n+1{)}^3$

Answer 1

The correct option is B $n^3$

Total number of terms upto $n^{th}$ row.
$1+2+3+4+...+n=\frac{n(n+1)}{2}$
i.e total number of $\frac{n(n+1)}{2}$ odd positive numbers are there upto $n^{th}$ row
Total number of terms upto $(n-1{)}^{th}$ row
$1+2+3+4+...+(n-1)=\frac{n(n-1)}{2}$
we know that sum of first $n$ odd positive numbers $=n^2$
Now sum of all odd positive numbers upto $n^{th}$ row is $S_n={\left(\frac{n(n+1)}{2}\right)}^2$
Similarly $S_{n-1}={\left(\frac{n(n-1)}{2}\right)}^2$
Sum of $n^{th}$ row = $S_n-S_{n-1}$
$={\left(\frac{n(n+1)}{2}\right)}^2-{\left(\frac{n(n-1)}{2}\right)}^2$
$=\frac{n^2}{4}\left[\right(n+1{)}^2-(n-1{)}^2]=\frac{n^2}{4}\left(4n\right)=n^3$