Algebra

Partial Fractions

Partial Fractions: Level 4 Challenges

         

11×2+11×5+13×3+12×7+15×4+ \frac {1}{1 \times 2} + \frac {1}{1 \times 5} + \frac {1}{3 \times 3} + \frac {1}{2 \times 7} + \frac {1}{5 \times 4} + \ldots

Find the sum of this infinite series.

S=11×2×3+12×3×4++1(n)(n+1)(n+2) S = \frac {1}{1\times 2 \times 3} + \frac {1}{2\times 3 \times 4} + \ldots + \frac {1} {(n)(n+1)(n+2)} ++114×(14+1)×(14+2)=ab,+ \ldots +\frac {1}{14 \times (14+1) \times (14+2)} = \frac {a} {b}, where aa and bb are positive, coprime integers. What is the value of a+b a + b ?

n=12015n2+n+1(n2+n)(n+1)!\displaystyle{\sum _{ n=1 }^{ 2015 }{ \frac { { n }^{ 2 }+n+1 }{ \left( { n }^{ 2 }+n \right) \left( n+1 \right) ! } }} can be represented in the form ab\dfrac { a }{ b } , where aa and bb are coprime positive integers. Find the value of aba-b.

n=0120152n2015(2n)\displaystyle\sum_{n=0}^{\infty} \dfrac{1}{2015^{2^{n}} - 2015^{-(2^{n})}}

If the closed form of the series above is in the form ab \frac a b , where aa and bb are positive coprime integers, then find ba.b - a.

For each positive integer nn, let Hn=11+12++1n.H_n = \frac{1}{1} + \frac{1}{2} + \cdots + \frac{1}{n}. If n=41nHnHn1=ab \sum_{n=4}^{\infty} \frac{1}{nH_nH_{n-1}} = \frac{a}{b} for relatively prime positive integers aa and bb, find a+ba+b.

This problem is shared by Sandeep S.

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