SAT1000 - P926

Let the $n$ th triangular number be $T(n) = \dfrac {n(n+1)}2$ . Then $b_1 = T(4)$ , $b_2 = T(5)$ , $b_3 = T(9)$ , $b_4 = T(10)$ , $b_5 = T(14)$ , $b_6 = T(15)$ , .... We have:

$b_n = \begin{cases} T\left(\frac {5(n-1)}2+4 \right) & \text{when } n \text{ is odd.} \\ T\left(\frac {5n}2\right) & \text{when } n \text{ is even.} \end{cases}$

Since $2k-1 = 4039$ is odd, we have

$b_{4039} = T \left(\frac {5(4039-1)}2 + 4 \right) = T(10099) = \frac {10099 \times 100100}2 = \boxed{50999950}$

Study the triangle numbers sequence modulo 5. You get
$1,3(1+2),1(1+2+3=6),0(1+2+3+4=10),0(0+5\equiv 0 \bmod 5)$ repeated over and over. This is equal to asking $Tri-num_{10099}.$