1 + 2 ! 1 2 + 2 2 + 3 ! 1 2 + 2 2 + 3 2 + 4 ! 1 2 + 2 2 + 3 2 + 4 2 + ⋯
If the value of the sum above can be expressed in the form of b a e , where a and b are coprime positive integers, find a + b .
Clarification:
e
≈
2
.
7
1
8
2
8
denotes the
Euler's number
.
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WOWWWW.BEAUTIFUL SOLUTION.
This is bashing. I have forgotten a little but it might be neatly done with exponential generating function. Maybe someone can help here.
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Yes... Using expansion series for e x and differentiating 3 times with appropriate manipulations gives n = 1 ∑ ∞ 6 n ! n ( n + 1 ) ( 2 n + 1 ) x n − 2 1 = 3 e x ( x 2 5 + 2 9 x 2 3 + 3 x 2 1 ) Just substituting x = 1 gives the result.
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Yeah maybe I was mistaken. That is not really "neat". I was thinking if there was a method to find
n = 1 ∑ ∞ k = 1 ∑ n k 2 n ! 1
as a probable product of two functions. Just like we are able to find n = 1 ∑ ∞ k = 1 ∑ n k 2 x n = 1 − x 1 n = 1 ∑ ∞ n 2 x n
I will check out.
Super cool dude. I have done this by making a series of summision in e. I just love your solution.
Nice solution!!
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We know that:
e x = n = 0 ∑ ∞ n ! x n
Differentiating with respect to x and multiplying by x three times one gets:
x e x = n = 0 ∑ ∞ n ! n x n
( x 2 + x ) e x = n = 0 ∑ ∞ n ! n 2 x n
( x 3 + 3 x 2 + x ) e x = n = 0 ∑ ∞ n ! n 3 x n
Making x = 1 :
e = n = 0 ∑ ∞ n ! n 2 e = n = 0 ∑ ∞ n ! n 2 5 e = n = 0 ∑ ∞ n ! n 3
Then:
S = n = 1 ∑ ∞ k = 1 ∑ n n ! k 2
S = 6 1 n = 1 ∑ ∞ n ! n ( n + 1 ) ( 2 n + 1 )
S = 6 1 n = 0 ∑ ∞ n ! n ( n + 1 ) ( 2 n + 1 )
S = 6 1 ⎣ ⎡ 2 n = 0 ∑ ∞ n ! n 3 + 3 n = 0 ∑ ∞ n ! n 2 + n = 0 ∑ ∞ n ! n ⎦ ⎤
S = 6 1 [ 2 ⋅ 5 e + 3 ⋅ 2 e + e ]
S = 6 1 7 e
Thus:
a = 1 7 , b = 6 , a + b = 2 3