a) lemma : $\frac{1}{k(k+1)}=\frac1k-\frac{1}{k+1}$ $\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\dots+\frac{1}{2014\cdot2015}=\frac11- \frac12+\frac12-...-\frac{1}{2015}=\frac11-\frac{1}{2015}=\frac{2014}{2015}$ so $m+n=2014+2015=4029$ b) We have to find $2017^{1444}\equiv?\pmod{2015}$ but $\phi(2015)=\phi(5)*\phi(13)*\phi(31)=4*12*30=1440$ so $2017^{1444}\equiv2^{1444}=2^4*2^{1440}=2^4=16\pmod{2015}$

$\sum_{i=1}^{2014}\frac{1}{i(i+1)}=\frac{2014}{2015}$ $m=2014$ and $n=2015$. $m+n=2014+2015=4029$ $m+3\equiv2^{1444}\hspace{1mm}(mod\hspace{1mm}2015)$ $\varphi(2015)=1440$ $2^{1444}\equiv2^4\equiv16\hspace{1mm}(mod\hspace{1mm}2015)$

Sadigly wrote: $\sum_{i=1}^{2014}\frac{1}{i(i+1)}=\frac{2014}{2015}$ $m=2014$ and $n=2015$. $m+n=2014+2015=4029$ $m+3\equiv2^{1444}\hspace{1mm}(mod\hspace{1mm}2015)$ $\varphi(2015)=1440$ $2^{1444}\equiv2^4\equiv16\hspace{1mm}(mod\hspace{1mm}2015)$ I solved similarly !! cool