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1 solution
n ⋅ ( n − 1 2 0 1 6 ) ( n 2 0 1 6 ) = n ⋅ ( n − 1 ) ! ( 2 0 1 7 − n ) ! 2 0 1 6 ! n ! ( 2 0 1 6 − n ) ! 2 0 1 6 ! = n ⋅ n ! ( 2 0 1 6 − n ) ! ( n − 1 ) ! ( 2 0 1 7 − n ) ! = n ⋅ n ( n − 1 ) ! ( 2 0 1 6 − n ) ! ( n − 1 ) ! ( 2 0 1 7 − n ) ( 2 0 1 6 − n ) ! = 2 0 1 7 − n ∴ n = 1 ∑ 2 0 1 6 n ( n − 1 2 0 1 6 ) ( n 2 0 1 6 ) = n = 1 ∑ 2 0 1 6 2 0 1 7 − n = 2 0 1 6 + 2 0 1 5 + … + 1 = 2 2 0 1 6 ⋅ 2 0 1 7 = 2 0 3 3 1 3 6 .
Note : ( b a ) = b ! ( a − b ) ! a !