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∑r=1n1r(r+1)(r+2)(r+3)=1a−1b(n+1)+1c(n+2)−1d(n+3)\sum _{r=1} ^n \frac{1}{r(r+1)(r+2)(r+3)} = \dfrac{1}{a} - \dfrac{1}{b(n+1)}+ \dfrac{1}{c(n+2)}- \dfrac{1}{d(n+3)}r=1∑nr(r+1)(r+2)(r+3)1=a1−b(n+1)1+c(n+2)1−d(n+3)1
If the equation above holds true for constants aaa, bbb, ccc and ddd, find a+b+c+da+b+c+da+b+c+d.
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