đáp án của biểu thức là gì hả bạn?
`2/42+2/56+2/72+............+2/(b(b+1))`
`=2(1/42+1/56+1/72+.............1/(b(b+1)))`
`=2(1/(6.7)+1/(7.8)+...........+1/(b(b+1)))`
`=2(1/6-1/7+1/7-1/8+.......+1/b-1/(b+1))`
`=2(1/6-1/(b+1))`
Đoạn nay bạn thay vào nhé!
Sửa đề: \(\dfrac{2}{62}\rightarrow\dfrac{2}{72}\)
Giải:
\(\dfrac{2}{42}+\dfrac{2}{56}+\dfrac{2}{72}+...+\dfrac{2}{b.\left(b+1\right)}\)
\(=\dfrac{2}{6.7}+\dfrac{2}{7.8}+\dfrac{2}{8.9}+...+\dfrac{2}{b.\left(b+1\right)}\)
\(=2.\left(\dfrac{1}{6.7}+\dfrac{1}{7.8}+\dfrac{1}{8.9}+...+\dfrac{1}{b.\left(b+1\right)}\right)\)
\(=2.\left(\dfrac{1}{6}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{8}+\dfrac{1}{8}-\dfrac{1}{9}+...+\dfrac{1}{b}-\dfrac{1}{b+1}\right)\)
\(=2.\left(\dfrac{1}{6}-\dfrac{1}{b+1}\right)\)
Giả sử, gọi:
\(2.\left(\dfrac{1}{6}-\dfrac{1}{b+1}\right)=a\)
\(\dfrac{1}{6}-\dfrac{1}{b+1}=a:2\)
\(\dfrac{1}{6}-\dfrac{1}{b+1}=\dfrac{a}{2}\)
\(\dfrac{1}{b+1}=\dfrac{1}{6}-\dfrac{a}{2}\)
\(\dfrac{1}{b+1}=\dfrac{1-3a}{6}\)
\(\Rightarrow\left(b+1\right).\left(1-3a\right)=1.6\)
\(\Rightarrow\left(b+1\right).\left(1-3a\right)=6\)
\(\Rightarrow b+1\) và \(1-3a\inƯ\left(6\right)=\left\{-6;-3;-2;-1;1;2;3;6\right\}\)
Vì b ∈ N nên b+1 và 1-3a ∈ Ư(6)={1;2;3;6}
Ta có bảng giá trị:
b+1=1 thì 1-3a=6
b=0 thì a=\(\dfrac{-5}{3}\)
b+1=6 thì 1-3a=1
b=5 thì a=0
b+1=2 thì 1-3a=3
b=1 thì a=\(\dfrac{-2}{3}\)
b+1=3 thì 1-3a=2
b=2 thì a=\(\dfrac{-1}{3}\)
Vậy b ∈ {0;1;2;5}
Chúc bạn học tốt!
