smc

The sum of the first $1$ numbers is $1$ The sum of the next $2$ numbers is $2+1$ The sum of the next $3$ numbers is $2+2+1$ In general, we can write "the sum of the next $n$ numbers is $1+2(n-1)$", where the word "next" follows the pattern established above. Thus, we first want to find what triangular numbers $1234$ is between. By plugging in various values of $n$ into $f(n)=\frac{n(n+1)}{2}$, we find: $f(50)=1275$ $f(49)=1225$ Thus, we want to add up all those sums from "next $1$ number" to the "next $49$ numbers", which will give us all the numbers up to and including the $1225^{th}$ number. Then, we can manually tack on the remaining $2$s to hit $1234$. We want to find: ${\sum }_{n=1}^{49}1+2(n-1)$ ${\sum }_{n=1}^{49}2n-1$ ${\sum }_{n=1}^{49}2n-{\sum }_{n=1}^{49}1$ $2{\sum }_{n=1}^{49}n-49$ $2\cdot \frac{49\cdot 50}{2}-49$ $49^2$ $2401$ Thus, the sum of the first $1225$ terms is $2401$. We have to add $9$ more $2$s to get to the $1234^{th}$ term, which gives us $2419$, or option $\boxed{2419}$. Note: If you notice that the above sums form $1+3+5+\mathrm{7...}+(2n-1)=n^2$, the fact that $49^2$ appears at the end should come as no surprise.