Healthy Algorithm

First of all we need to determine the first number of the ${45}^{th}$ row. The sequence of first numbers of each row is: $1,2,4,7,11 . .$ We can observe that the second order difference i.e. difference of differences is a constant value.
So lets consider a quadratic expression $P(n):a{ n }^{ 2 }+bn+c$
$P(1):a+b+c=1\\P(2):4a+2b+c=2\\P(3):9a+3b+c=4$
Solving the system of equations, we obtain $a=\frac{1}{2},b=\frac{-1}{2},c=1$ $\Rightarrow P(n):\frac{1}{2}{ n }^{ 2 }-\frac{1}{2}n+1\\\Rightarrow P(45)=991$ .
The last number of the ${45}^{th}$ row will be $991+44=1035$ .
Hence the ${45}^{th}$ row is: ${991}^{3},{992}^{3},{993}^{3} . . . {1035}^{3}\\=\displaystyle\sum _{ r=1 }^{ 1035 }{ { r }^{ 3 } }-\displaystyle\sum _{ r=1 }^{ 990 }{ { r }^{ 3 } }\\={ \left( \frac { 1035\times 1036 }{ 2 } \right) }^{ 2 }-{ \left( \frac { 991\times 992 }{ 2 } \right) }^{ 2 }\\=287435376900-240634397025\\\Huge\color{#3D99F6}{=\boxed{46800979875}}$