Grouping Odds

Let the sum of $n$ th group be $S_n$ . We note that $S_1=1$ , $S_2=8=2^3$ , $S_3=27=3^3$ .... It seems to imply that $S_n = n^3$ . Let us prove it.

Let the terms be $a_k$ and the number of terms in a group be $m$ , so that:

$\underbrace{\left \{ a_1 \right \}}_{m=1}, \underbrace{\left \{ a_2, a_3 \right \}}_{m=2}, \underbrace{\left \{ a_4, a_5, a_6 \right \}}_{m=3} ..., \underbrace{\left \{ a_{\frac {(n-2)(n-1)}2+1}, ..., a_{\frac {(n-1)n}2} \right \}}_{m=n-1}, \underbrace{\left \{ a_{\frac {(n-1)n}2+1}, ..., a_{\frac {n(n+1)}2} \right \}}_{m=n}$

We note that the $m$ th group has $m$ terms, and that the index ( $k$ ) of the last term of $m$ th group is a triangular number $k = \dfrac {m(m+1)}2$ , so that $m = 1,2,3..., n$ $\implies k = 1,3,6..., \dfrac {n(n+1)}2$ . And the index of the first term of $m$ th group is $k=\dfrac {(m-1)m}2+1$ . We also note that $a_k = 2k-1$ . Therefore, the sum of $n$ th ( $m=n$ ) group is given by:

$\begin{aligned} S_n & = \frac {N(A+L)}2 & \small \color{#3D99F6} \text{where }N \text{ is the number of terms}, A, L, \text{the first and last terms.} \\ & = \frac n2 \left(a_{\frac {(n-1)n}2+1} + a_{\frac {n(n+1)}2} \right) \\ & = \frac n2 \left( 2 \left(\frac {(n-1)n}2+1 \right) -1 + 2\left(\frac {n(n+1)}2 \right)-1 \right) \\ & = \frac n2 \left( (n-1)n+n(n+1) \right) \\ & = n^3 \end{aligned}$

$\implies S_{10} = 10^3 = \boxed{1000}$ .

1 + 2 + 3 + 4 + ..... + 10 = 55. Since 1st group has 1 term , 2nd group has 2 terms , ... ,

nth group has n terms. There are 55 terms upto tenth group ( including tenth group)

$55^2$ is the sum of the numbers upto last term in the tenth group.

1 + 2 + 3 + ....+ 9 = 45 , similarly there are 45 terms upto ninth group (including ninth group)

$45^2$ is the sum of the numbers upto last term in the ninth group.

So therefore, sum of numbers present in the tenth group will be

$55^2$ - $45^2$ = (55+45)(55-45) = (100)(10) = 1000.

USED RESULTS :

1) sum of first n natural numbers = n(n+1)/2 .

2) sum of first n odd numbers = $n^2$ .

If we compute the sum in the first groups, we obtain $1, 8, 27, 64$ . We can notice that the sum in the $i^{th}$ group is $i^3$ .

Thus, in the tenth group the sum of the elements is $10^3=1000$ .