1 2 + ( 1 2 + 2 2 ) + ( 1 2 + 2 2 + 3 2 ) + … + ( 1 2 + … + 1 9 9 2 + 2 0 0 2 ) = ?
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Two equivalent ways to obtain the result:
E x p r e s s i o n = 2 0 0 ∗ 1 2 + 1 9 9 ∗ 2 2 + . . . + 1 ∗ 2 0 0 2 , = i = 1 ∑ n ( 2 0 1 − i ) ∗ i 2 = i = 1 ∑ n ( 2 1 0 ∗ i 2 − i 3 ) = 2 0 1 ∗ 6 n ∗ ( n + 1 ) ∗ ( 2 n + 1 ) − ( 2 n ∗ ( n + 1 ) ) 2 = 2 n ∗ ( n + 1 ) ∗ { 2 0 1 ∗ 3 2 n + 1 − 2 n ∗ ( n + 1 ) } B u t n = 2 0 0 , ∴ E x p r e s s i o n = 2 2 0 0 ∗ 2 0 1 ∗ { 3 4 0 1 ∗ 2 0 1 − 2 2 0 0 ∗ 2 0 1 } = 1 3 6 0 1 6 7 0 0 .
The basics
Then figure it out. Sorry for the handmade
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The above expression can be represented by n = 1 ∑ 2 0 0 k = 1 ∑ n k 2 .
And k = 1 ∑ n k 2 = 6 n ( n + 1 ) ( 2 n + 1 ) .
Thus,
n = 1 ∑ 2 0 0 k = 1 ∑ n k 2 = = = = = = = n = 1 ∑ 2 0 0 6 n ( n + 1 ) ( 2 n + 1 ) n = 1 ∑ 2 0 0 6 2 n 3 + 3 n 2 + n n = 1 ∑ 2 0 0 ( 3 1 n 3 + 2 1 n 2 + 6 1 n ) 3 1 n = 1 ∑ 2 0 0 n 3 + 2 1 n = 1 ∑ 2 0 0 n 2 + 6 1 n = 1 ∑ 2 0 0 n 3 1 ( 2 2 0 0 ( 2 0 1 ) ) 2 + 2 1 ( 6 2 0 0 ( 2 0 1 ) ( 4 0 1 ) ) + 6 1 ( 2 2 0 0 ( 2 0 1 ) ) 3 1 ( 4 0 4 0 1 0 0 0 0 ) + 2 1 ( 2 6 8 6 7 0 0 ) + 6 1 ( 2 0 1 0 0 ) 1 3 4 6 7 0 0 0 0 + 1 3 4 3 3 5 0 + 3 3 5 0 = 1 3 6 0 1 6 7 0 0