Question

If odd natural numbers are arranged in groups as $\left(1\right),(3,5),(7,9,11),...$ Then the sum of the numbers in the ${10}^{\text{th}}$ group is

• A. $500$
• B. $729$
• C. $743$
• D. $1000$

Answer 1

The correct option is D $1000$

$n^{\text{th}}$ group will have ${}^{\prime }{n}^{\prime }$ natural odd numbers
Consider first numbers in groups $1,3,7,13$
Let $a_1=1,a_2=3,a_3=7,a_4=13$ and so on
$\Delta a_1=2,4,6,..$ are in A.P.
(where $\Delta a_1$ is first difference of the terms)
$\therefore a_n=1+(2+4+6+...\text{upto}(n-1\left)\text{terms}\right)$
$=1+n(n-1)=n^2-n+1$
$\therefore n^{\text{th}}$ group will have $n^2-n+1,n^2-n+3,...\text{upto}n\text{terms}$
Sum $=n(n^2-n)+(1+3+5+...\text{upto}n\text{terms})=n^3-n^2+n^2=n^3$
For ${10}^{\text{th}}$ group
Sum $={10}^3=1000$

Alternate Solution
the sum of the numbers in first group
$1=1^3$
the sum of the numbers in $2^{\text{nd}}$ group
$3+5=8=2^3$
the sum of the numbers in $3^{\text{rd}}$ group
$7+9+11=27=3^3$
So, the sum of the numbers in ${10}^{\text{th}}$ group
is ${10}^3=1000$