Do you play hockey?

Here's a novel but refreshingly nice way to solve this problem.

First, note that:

$\begin{aligned} T_{n} = 1 + 2 + \dots + n = \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \end{aligned}$

Given that:

$\begin{aligned} 1^2 + 2^2 + \dots + n^2 = \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \end{aligned}$

then:

$\begin{aligned} \sum_{k=1}^{n} T_{k} &= \sum_{k=1}^{n} \frac{k(k+1)}{2} \\ &= \frac{1}{2} \sum_{k=1}^{n} k^2 + \frac{1}{2} \sum_{k=1}^{n} k \\ &= \frac{n(n+1)(2n+1)}{12} + \frac{n(n+1)}{4} \\ &= \frac{n(n+1)(2n+1)}{12} + \frac{3n(n+1)}{12} \\ &= \frac{n(n+1)(2n+1+3)}{12} \\ &= \frac{n(n+1)(n+2)}{6} \end{aligned}$

Let:

$\begin{aligned} S_n = \frac{T_1 + T_2 + \dots + T_n}{1+2+\dots+n} \end{aligned}$

So:

$\begin{aligned} S_n &= \frac{\frac{n(n+1)(n+2)}{6}}{\frac{n(n+1)}{2}} \\ &= \frac{n+2}{3} \end{aligned}$

and:

$\begin{aligned} S_{1234} = \frac{1236}{3} = 412 \end{aligned}$

as required.

Relevant wiki: Sum of n, n², or n³

$\begin{aligned} Q_n & = \frac {T_1+T_2+T_3+...+T_n}{1+2+3+...+n} \\ & = \frac {n+2(n-1)+3(n-2)+...+n}{1+2+3+...n} \\ & = \frac {\sum_{k=1}^n k(n+1-k)}{\sum_{k=1}^n k} \\ & = \frac {(n+1)\sum_{k=1}^n k - \sum_{k=1}^n k^2}{\sum_{k=1}^n k} \\ & = n+1 - \frac {\sum_{k=1}^n k^2}{\sum_{k=1}^n k} \\ & = n+1 - \frac {\frac {n(n+1)(2n+1)}6}{\frac {n(n+1)}2} \\ & = n+1 - \frac {2n+1}3 \\ & = \frac {n+2}3 \end{aligned}$

$\begin{aligned} \implies Q_{1234} & = \frac {1234+2}3 = \boxed{412} \end{aligned}$

$\text {Let's solve the problem in general (for } n \in \mathbb {N} , n > 1 \text {) first: }$

$T_n = 1+2+...+n = \frac {n(n+1)}{2} = 0.5(n^2 + n)$

$\frac {T_1+T_2+...+T_n}{1+2+...+n} = \frac { 0.5 ((1^2 + 2^2 + ... + n^2) + (1+2+...+n)) }{1+2+...+n} =$

$= \frac { 0.5 ( \frac {n(n+1)(2n+1)}{6}+ 0.5n(n+1)) }{0.5n(n+1)} = \frac { n(n+1)( \frac {2n +1}{6} + 0.5) }{n(n+1)} = \frac {2n +1}{6} + 0.5 =\frac {2n +4}{6}$

In our case, n = 1234 :

$\frac {2n +4}{6} = \frac {2 × 1234 +4}{6} = \frac {2472}{6} = \boxed {412}$

Tetrahedral number / triangular number