Adding with cousins

Number Theory Level 3

For all $n\ge 2$ , there are ${ p }_{ n }$ the prime number less than or equal to $n$ and ${ q }_{ n }$ the next prime number larger than $n$ . For example: $n=3$ , ${ p }_{ 3 }=3$ and ${ q }_{ 3 }=5$ .

Find the value of

$\frac { 1 }{ { p }_{ 2 }{ q }_{ 2 } } +\frac { 1 }{ { p }_{ 3 }{ q }_{ 3 } } +\frac { 1 }{ { p }_{ 4 }{ q }_{ 4 } } +\frac { 1 }{ { p }_{ 5 }{ q }_{ 5 } } +\frac { 1 }{ { p }_{ 6 }{ q }_{ 6 } } +\frac { 1 }{ { p }_{ 7 }{ q }_{ 7 } } +\frac { 1 }{ { p }_{ 8 }{ q }_{ 8 } } +\frac { 1 }{ { p }_{ 9 }{ q }_{ 9 } } +\frac { 1 }{ { p }_{ 10 }{ q }_{ 10 } }$

\frac 1{77}

\frac 9{22}

\frac 12

\frac {28}{2\cdot3\cdot5\cdot7\cdot11}

\frac {10}{11}

2 solutions

Md Zuhair
Apr 13, 2017

So putting $n=10$ , we get $\dfrac{9}{22}$

How do you get the sum ?

Kushal Bose - 4 years, 1 month ago

Brian Moehring
Aug 22, 2018

Let $p(n)$ denote the $n$ th prime number. Then $p(5)=11,$ so $\begin{aligned} \sum_{k=2}^{11-1} \frac{1}{p_kq_k} &= \sum_{m=1}^{5-1}\quad \sum_{k=p(m)}^{p(m+1)-1} \frac{1}{p(m)p(m+1)} \\ &= \sum_{m=1}^{5-1} \frac{p(m+1)-p(m)}{p(m)p(m+1)} \\ &= \sum_{m=1}^{5-1} \frac{1}{p(m)} - \frac{1}{p(m+1)} \\ &= \frac{1}{p(1)} - \frac{1}{p(5)} \\ &= \frac{1}{2} - \frac{1}{11} \\ &= \boxed{\frac{9}{22}} \end{aligned}$

Adding with cousins

2 solutions

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