Question

determinați n aparține lui N pentru care 1/1×2+1/2×3+1/3×4+1/4×5+.....+1/n(n+1)<6/7​

Answer 1

Răspuns:

Explicație pas cu pas:

$\dfrac{1}{k(k+1)}=\dfrac{(k+1)-k}{k(k+1)}=\dfrac{k+1}{k(k+1)}-\dfrac{k}{k(k+1)}=\dfrac{1}{k}-\dfrac{1}{k+1}\\\texttt{Suma poate fi telescopata : }\\\dfrac{1}{1\cdot 2}+\dfrac{1}{2\cdot 3}+\dfrac{1}{3\cdot 4}+\ldots+\dfrac{1}{n(n+1)}<\dfrac{6}{7}\\1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\ldots+\dfrac{1}{n}-\dfrac{1}{n+1}<\dfrac{6}{7}\\\texttt{Termenii se simplifica si at the end of the day mai ramane:}\\1-\dfrac{1}{n+1}<\dfrac{6}{7}\\\dfrac{n}{n+1}<\dfrac{6}{7}\\7n<6n+6$

$n<6\Rightarrow n\in \{1,2,3,4,5\}$