n = 1 ∑ ∞ n 2 ( n + 1 ) ( n + 2 ) 1 = b π a − d c
The above sum is true for integers a , b , c , d , where c , d are coprime positive integers. Give your answer as a + b + c + d .
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Brilliant ,just brilliant Mr.Alak .
S = n = 1 ∑ ∞ n 2 ( n + 1 ) ( n + 2 ) 1 = n = 1 ∑ ∞ ( n ( n + 1 ) 1 ) ( n ( n + 2 ) 1 ) = n = 1 ∑ ∞ ( n 1 − n + 1 1 ) ⋅ 2 1 ( n 1 − n + 2 1 ) = n = 1 ∑ ∞ 2 1 ( n 2 1 − n ( n + 1 ) 1 − n ( n + 2 ) 1 + ( n + 1 ) ( n + 2 ) 1 ) = n = 1 ∑ ∞ 2 1 ( n 2 1 − n 1 + n + 1 1 − 2 1 ( n 1 − n + 2 1 ) + n + 1 1 − n + 2 1 ) = n = 1 ∑ ∞ 2 1 ( n 2 1 − 2 n 3 + n + 1 2 − 2 ( n + 2 ) 1 ) = 2 1 n = 1 ∑ ∞ n 2 1 − 4 3 n = 1 ∑ ∞ ( n 1 − n + 1 1 ) + 4 1 n = 1 ∑ ∞ ( n + 1 1 − n + 2 1 ) = 2 1 ζ ( 2 ) − 4 3 ⋅ 1 + 4 1 ⋅ 2 1 = 1 2 π 2 − 8 5 Riemann zeta function ζ ( s ) = k = 1 ∑ ∞ k s 1 and ζ ( 2 ) = 6 π 2
Therefore a + b + c + d = 2 + 1 2 + 5 + 8 = 2 7 .
Reference: Riemann zeta function
Another incredible solution,but there is a problem with zeta of 2 ,sir you are forgetting π 2
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n 2 ( n + 1 ) ( n + 2 ) 1 = 2 n 2 1 − 4 3 ( n 1 − n + 1 1 ) + 4 1 ( n + 1 1 − n + 2 1 )
Therefore the sum is equal to 2 1 × 6 π 2 − 4 3 × 1 + 4 1 × 2 1 = 1 2 π 2 − 8 5
Hence a = 2 , b = 1 2 , c = 5 , d = 8 , and a + b + c + d = 2 7 .