2018 AMC 12B Problem #9

$\begin{aligned} S & = \sum_{i=1}^n \sum_{j=1}^n (i+j) \\ & = \sum_{i=1}^n \left(ni+\frac {n(n+1)}2\right) \\ & = n\times \frac {n(n+1)}2 + \frac {n(n+1)}2\times n \\ & = n^2(n+1) \quad \quad \small \color{#3D99F6} \text{Putting }n=100 \\ & = \boxed{1010000} \end{aligned}$

$=\quad \sum _{ i=1 }^{ 100 }{ \sum _{ j=1 }^{ 100 }{ (i+j) } } \\ =\quad \sum _{ i=1 }^{ 100 }{ (i\quad +\quad \frac { 100(100+1) }{ 2 } ) } \\ =\quad \sum _{ i=1 }^{ 100 }{ (i\quad +\quad 5050) } \\ =\quad \sum _{ i=1 }^{ 100 }{ i } +\sum _{ i=1 }^{ 100 }{ 5050 } \\ =\quad 5050\quad +\quad 505000\\ =\quad 1010000\\$