Does cover up rule help?

Algebra Level 4

The partial fraction of 1 ( x + 1 ) ( x + 2 ) ( x + 3 ) ⋯ ( x + 100 ) \dfrac1{(x+1)(x+2)(x+3)\cdots(x+100) } can be written as ∑ j = 1 100 a j x + j \displaystyle \sum_{j=1}^{100} \dfrac{a_j}{x+j} , where a 1 , a 2 , … , a 100 a_1,a_2,\ldots,a_{100} are constants. What is the value of a 1 + a 2 + ⋯ + a 100 a_1 + a_2 + \cdots + a_{100} ?


The answer is 0.

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Pi Han Goh by Pi Han Goh , 30, Malaysia

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1 solution

Aaron Jerry Ninan
Jul 29, 2016

Take partial fraction of each 2 consecutive term. it is seen that a1=1 a2=-1 this pattern goes for every 2 consecutive term upto hundred .all the +1 and -1 pairs get cancelled. therfore answer is 0 .

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