a) [tex]\frac{1}{n(n+1)}=\frac{(n+1)-n}{n(n+1)}=\frac{n+1}{n(n+1)}-\frac{n}{n(n+1)}=\frac{1}{n}-\frac{1}{n+1}[/tex]
b) [tex]\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{n(n+1)}=\\ \\ =\frac{1}{2\cdot3} +\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+...+\frac{1}{n(n+1)}=\\ \\=(\frac{1}{2}-\frac{1}{3})+(\frac{1}{3}-\frac{1}{4})+(\frac{1}{4}-\frac{1}{5})+...+(\frac{1}{n}-\frac{1}{n+1}); \ (din \ punctul \ a \ avem \ scrierea \ aceasta)\\ \\ =\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{n}-\frac{1}{n+1}=\\ \\=\frac{1}{2}-\frac{1}{n+1}; \ (restul \ termenilor \ se \ reduc)\\ \\ =\frac{n+1}{2(n+1)}-\frac{2}{2(n+1)}=\\ \\=\frac{n+1-2}{2(n+1)}=\\ \\=\frac{n-1}{2(n+1)}\\ \\Dar \ N=\frac{999}{2000} \Rightarrow \frac{n-1}{2(n+1)}=\frac{999}{2000} \Rightarrow 2000(n-1)=999(2n+2) \\ \\ \Rightarrow 2000n-2000=1998n+1998 \Rightarrow 2000n-1998n=2000+1998 \Rightarrow 2n=3998 \\ \\ \Rightarrow \boxed{n=1999}[/tex]
Explicație pas cu pas:
Se considera numarul real N=1/6 + 1/12 + 1/20 +.....+ 1/n(n+1) a) Aratati ca 1/n(n+1)=1/n - 1/n+1 b) Determinati n astfel incat N=999/2000
a)1/n(n+1)=(n+1-n)/n(n+1)=(n+1)/n(n+1)-n/n(n+1)=1/n-1/(n+1)
b)
1/6 + 1/12 + 1/20 +.....+ 1/n(n+1) =999/2000
1/2*3+1/4*3+1/4*5+...+1/n(n+1)=999/2000
(3-2)/2*3+(4-3)/4*3+(5-4)/4*5+...+(n+1-n)/n(n+1)=999/2000
3/(2*3)-2/(2*3)+4/(4*3)-3/(4*3)+5/(4*5)-4/(4*5)+...+(n+1)/n(n+1)-n/n(n+1)=999/2000
1/2-1/3+1/3-1/4+1/4-1/5+...1/n-1/(n+1)=999/2000
1/2-1/(n+1)=999/2000 |*2000(n+1)
1000(n+1)-2000=999(n+1)
1000(n+1)-999(n+1)=2000
(n+1)(1000-999)=2000
n+1=2000
x=2000-1
n=1999
Bafta!
