Happy Victory Day, Vietnam!
Find 1 9 7 5 ! f ( 1 9 7 5 ) ( 0 ) for f ( x ) = ( 1 − x ) 4 ( x + 1 ) x .
Clarifications :
- f ( k ) ( x ) denotes the k th derivative of f ( x ) .
- ! denotes the factorial notation. For example, 8 ! = 1 × 2 × 3 × ⋯ × 8 .
The answer is 2569862100.
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3 solutions
Yes, exactly! (+1) Note that the answer is ∑ k = 1 1 9 7 5 k 2 . In fact, this is one way to derive the formula ∑ k = 1 n k 2 = 6 n ( n + 1 ) ( 2 n + 1 )
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can you please tell how this proves that?
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We have the generating function ∑ n = 1 ∞ n 2 x n = ( 1 − x ) 3 x ( x + 1 ) . Multiplying through with ∑ n = 0 ∞ x n = 1 − x 1 gives ∑ n = 1 ∞ ( ∑ k = 1 n k 2 ) x n = ( 1 − x ) 4 x ( x + 1 )
Relevant wiki: Generating Functions
If A ( x ) is the generating function of { a n } , then 1 − x A ( x ) is the generating function of the partial sums of a n . Now, A ( x ) = x + x 2 is the generating function of the sequence { 0 , 1 , 1 , 0 , . . . , 0 , . . . } . Denote by { S a n i } the partial sums of the sequence { S a n i − 1 } , and define { S a n 0 } = { a n } . We have: S a n 1 = { 0 , 1 , 2 , 2 , . . . , 2 , . . . } and: S a n 2 = { 0 , 1 , 3 , 5 , 7 , . . . , 2 n − 1 , . . . } and: S a n 3 = { 0 , 1 , 4 , 9 , . . . , n 2 , . . . } and finally: S a n 4 = { 0 , 1 , 5 , 1 4 , 3 0 , . . . , 6 n ( n + 1 ) ( 2 n + 1 ) , . . . } . The 1975th element in this sequence is the answer.
Yes, exactly! (+1) That's the solution I had in mind! People who know about generating functions would probably be familiar with the generating function of n 2 , so, in my comment to Abhi's solution I took it from there. Then we take the generating function of the partial sums once more to find the answer.
