Question

$\text{Calc. suma:}~ \sqrt{1+ \frac{1}{ 1^{2} }+ \frac{1}{ 2^{2} }}+\sqrt{1+ \frac{1}{ 2^{2} }+ \frac{1}{ 3^{2} }}+...+\sqrt{1+ \frac{1}{ 2013^{2} }+ \frac{1}{ 2014^{2} }}.$

Answer 1

$Avem:~1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2} =calcule= \frac{k^4+k^2+1+2k^3+2k^2+2k}{k^2(x+1)^2} = \\ \\ = \frac{(k^2+k+1)^2}{k^2(k+1)^2}. \\ \\ Cum~se~observa~ca~k^4+k^2+1+2k^3+2k^2+2k=(k^2+k+1)^2~? \\ \\ Se~"intuieste"~ca~ar~fi~un~patrat,~si,~in~mod~evident~nu~patratul~al~ \\ \\ unei~sume~de~2~termeni,~ci~de~3.~"k^4"~ne~indica~ca~in~scrierea \\ \\ (a+b+c)^2,~a=k^2,~iar~"+1"~necesita~c=1.~"2k^3=2 \cdot k^2 \cdot k \Rightarrow \\ \\b=k.$

$Revenind:~ \sqrt{1+ \frac{1}{k^2}+ \frac{1}{(k+1)^2} }= \frac{k^2+k+1}{k(k+1)}= \frac{k}{k+1}+ \frac{1}{k}=1- \frac{1}{k+1}+ \frac{1}{k}= \\ \\ S= (1+ \frac{1}{1}- \frac{1}{2})+(1+ \frac{1}{2}- \frac{1}{3})+(1+ \frac{1}{3}- \frac{1}{4})+...+(1+ \frac{1}{2013}- \frac{1}{2014})= \\ \\ =2013+ \frac{1}{1}- \frac{1}{2014}= \\ \\ = \boxed{2014- \frac{1}{2014} } .$