Given that ${\sum\limits_{r=0}^{N}{\frac{1}{\left( r+1 \right)\left( r+3 \right)}}=\frac{3}{4}-\frac

Question

Given that ${\sum\limits_{r=0}^{N}{\frac{1}{\left( r+1 \right)\left( r+3 \right)}}=\frac{3}{4}-\frac{1}{2\left( N+2 \right)}-\frac{1}{2\left( N+3 \right)}}$, find ${\sum\limits_{r=4}^{N+3}{\frac{1}{\left( r-1 \right)\left( r-3 \right)}}}$ ${\sum\limits_{r=5}^{N}{\frac{1}{\left( r-1 \right)\left( r-3 \right)}}}$

Accepted Answer

(i) 3/4 - 1/(2(N+1)) - 1/(2(N+2)), (ii) 5/12 - 1/(2(N-2)) - 1/(2(N-1))

6.6 Sigma Notation By Substitution Q1

6.6 Sigma Notation By Substitution Q1

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