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Ajutor ! Determinati K astfel incat [tex] \frac{2+k}{1} + \frac{3+k}{2} + \frac{4+k}{3} +....+ \frac{n+k}{n-1} = n-1 [/tex] , pentru oricare n [tex] \geq [/tex] 2 numar natural.

Răspuns :

fiecare termen il descompui:
(2+k)/1 =(1+(1+k)/1)
(3+k)/2=(1+(1+k)/2)
..................................
(n+k)/(n-1)=( 1+(1+k)/(n-1) )
notez suma termenilor cu S
si cand le aduni da:
S = (1+(1+k)/1)+(1+(1+k)/2)+...+( 1+(1+k)/(n-1) )
ai (n-1) paranteze, fiecare continandu-l pe 1 =>
=> S =(n-1) +[ (1+k)/1 +(1+k)/2 +...+(1+k)/(n-1)]
dar S=(n-1) =>
=> (n-1) +[ (1+k)/1 +(1+k)/2 +...+(1+k)/(n-1)] =(n-1) <=>
<=>(1+k)/1 +(1+k)/2 +...+(1+k)/(n-1) =0 <=>
<=>(1+k) [1/1+1/2+1/3+...+1/(n-1) ]=0
<=> (1+k) =0 <=>
<=>k = -1