AIME Problem

Note ${ S }_{ n }= \sum _{ k=1 }^{ n }{ \frac { 1 }{ { k }^{ 2 }+k } } =\sum _{ k=1 }^{ n }{ \frac { 1 }{ k } -\frac { 1 }{ (k+1) } }$ . Hence applying ${ V }_{ n }$ method we get ${ S }_{ n }= 1-\frac { 1 }{ 1+n }$ . Now we had ${ a }_{ m }+{ a }_{ m+1 }........+{ a }_{ n-1 }=\frac { 1 }{ 31 }$ . Now ${ a }_{ m }+{ a }_{ m+1 }........+{ a }_{ n-1 }=({ a }_{ 1 }+{ a }_{ 2 }+{ a }_{ 3 }........+{ a }_{ n-1 })-({ a }_{ 1 }+{ a }_{ 2 }+.......+{ a }_{ m-1 }) = { S }_{ n-1 }-{ S }_{ m-1. }$

Applying our result we get $\frac { 1 }{ 31 } = (1-\frac { 1 }{ n } )-(1-\frac { 1 }{ m } )= \frac { 1 }{ m } -\frac { 1 }{ n }$ . Rearranging terms we get $31(n-m)=mn$ . Now since $31$ is a prime implies either of $m$ or $n$ is a multiple of $31$ . Let's assume $m=31a$ where $a \epsilon I$ . Then $n-31a=an \Rightarrow n = \frac { 31a }{ 1-a }$ But since $a\ge 1$ implies $n$ is negative so $m \neq 32a$ . Let's assume $n = 31a$ where $a \epsilon I$ Then $m = \frac { 31a }{ a+1 }$ . Since $m$ is an integer implies $a=30$ . Putting $a=30$ we get $m=30$ $n=930$ . So $m+n=960$

$a_k=\frac{1}{k^2+k}=\frac{1}{k}-\frac{1}{k+1}$ , so $a_m+a_{m+1}+...+a_{n-1}=\frac{1}{m}-\frac{1}{m+1}+\frac{1}{m+1}-\frac{1}{m+2}+...+\frac{1}{n-1}-\frac{1}{n}=\frac{1}{m}-\frac{1}{n}$ .

$\frac{1}{m}-\frac{1}{n}=\frac{1}{31}$

$31n-31m-mn=0$

$(n+31)(31-m)=961$

Since $n$ must be positive, so is $n+31$ . There are two ways to multiply two positive integers to get $961$ : $1*961$ and $31*31$ .

$n+31$ cannot equal neither $1$ nor $31$ if $n$ is positive, so we have $n+31=961$ and $31-m=1$ , so $m=30$ , $n=930$ , and $m+n=\fbox{960}$ .

1/k^2 + k is 1/k - 1/k+1. Out of all numbers n less than 30 plugged into 1/n - x = 1/31, only 30 yields 1 on the numerator for x. 30 + 930 (which is the denominator of 1/n-1 + 1) = 960