## Sum of Odd Numbers

Suppose the odd numbers are grouped in the following way: $\{1\},\{3,5\},\{7,9,11\},\{13, 15, 17, 19\},\cdots$ What is the sum of the numbers in the tenth grouping?
Source: NCTM Mathematics Teacher 2006

SOLUTION
Brute force method
It doesn’t take long to write out the $60$ consecutive odd numbers $1,3,5,7,\cdots,119$ and notice that the tenth group consists of $\{91,93,95,97,99,101,103,105,107,109\}$. The sum of the numbers in the tenth grouping equals $1000$.

Pattern method
Group $2$ has $2$ elements $\{3,5\}$
Value of first element $=2(2-1)+1=3$
Value of last element $=3+(2-1)2=5$
Sum of elements $=3+5=8=2^3$

Group $3$ has $3$ elements $\{7,9,11\}$
Value of first element $=3(3-1)+1=7$
Value of last element $=7+(3-1)2=11$
Sum of elements $=7+9+11=27=3^3$

Group $4$ has $4$ elements $\{13,15,17,19\}$
Value of first element $=4(4-1)+1=13$
Value of last element $=13+(4-1)2=19$
Sum of elements $=13+15+17+19=64=4^3$

In general group $i$ has $i$ elements
Value of first element $=i(i-1)+1$
Value of last element $=i(i-1)+1+(i-1)2=(i-1)(i+2)+1$
We prove that the sum of elements $=i^3$.
Recall that given $n$ consecutive odd numbers $\{a_1,a_2,a_3,\cdots,a_n\}$
$a_1+a_2+a_3+\cdots+a_n=\dfrac{(a_1+a_n)}{2}\times n$
Sum of elements of group $i=\dfrac{i(i-1)+1+(i-1)(i+2)+1}{2}\times i$
$=\dfrac{i^2-i+1+i^2+i-1}{2}\times i$
$=\dfrac{2i^2}{2}\times i$
$=i^3$
Sum of the numbers in the tenth grouping
$10^3=1000$

Answer: $1000$