las 1 deoparte
1/(1+2) =2/2*3
1/(1+2+3)=2/3*4
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1/(1+...+x)=2/x*(x+1)
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1/(1+2) +...+1/(1+2+...+x) =2/2*3+2/3*4+...+2/x*(x+1)=
=2*(1/2*3+1/3*4+...+1/x*(x+1)) (ceea ce este in paranteza mai are si denumirea de suma telescopica( atunci cand la numitor ai produs de 2 numere, unul din numere repetandu-se la urmatoarea fractie)
1/2*3 =1/2 -1/3
1/3*4=1/3-1/4
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1/x*(x+1)=1/x-1/(x+1)
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1/2*3+...+1/x*(x+1)=1/2-1/3+1/3-1/4+...+1/x -1/(x+1) (observi ca termenii se reduc, ramane : )
1/2-1/(x+1)
cum noi aveam de calculat 2*(1/2*3+1/3*4+...+1/x*(x+1)) =>
=> 2*[ 1/2-1/(x+1) ] = 1-2/(x+1) , iar la final adun si 1 ce l-am lasat deoparte => 2-2/(x+1)=200/101 <=>
<=>2x/(x+1)=200/101 <=> x=100
