Sao un dài thế? Hay là Sn? Chắc là Sn đó

\(C^3_n=\dfrac{n!}{3!.\left(n-3\right)!}=\dfrac{n\left(n-1\right)\left(n-2\right)}{6}\)

\(\Rightarrow\dfrac{1}{C^3_n}=\dfrac{6}{n\left(n-1\right)\left(n-2\right)}\)

\(\Rightarrow S_n=\dfrac{6}{1.2.3}+\dfrac{6}{2.3.4}+\dfrac{6}{3.4.5}+\dfrac{6}{4.5.6}+...+\dfrac{6}{n\left(n-1\right)\left(n-2\right)}\)

Này hình như toán lớp 6 thì phải, chả nhớ :v

\(\dfrac{1}{n\left(n-1\right)\left(n-2\right)}=\dfrac{n-\left(n-2\right)}{2.n\left(n-1\right)\left(n-2\right)}=\dfrac{1}{2\left(n-1\right)\left(n-2\right)}-\dfrac{1}{2n\left(n-1\right)}=\dfrac{1}{2}\left(\dfrac{1}{n-1}.\dfrac{1}{n-2}-\dfrac{1}{n-1}.\dfrac{1}{n}\right)\)

\(\dfrac{1}{1.2.3}=\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{6}\right);\dfrac{1}{2.3.4}=\dfrac{1}{2}\left(\dfrac{1}{6}-\dfrac{1}{12}\right);...\)

Cộng lại thì sẽ triệt tiêu mấy phần tử 1/6; 1/12;...

\(\Rightarrow S_n=6.\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n-1\right)}\right)=3\left(\dfrac{1}{2}-\dfrac{1}{n\left(n-1\right)}\right)\)

\(\Rightarrow lim\left(\dfrac{3}{2}-\dfrac{3}{n^2-n}\right)=\dfrac{3}{2}\)

Lâu ko ôn lại cũng miss cách tính limit luôn :v Cơ mà có khi bằng 3/2 thiệt á, check lại hộ tui xem

Ai là hs giỏi thì giúp mik vs mai nộp rồi

Ta có : \(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy a = 1,b = 2,c = 3,d = 4

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy a = 1;b=2;c=3;d=4 ( thỏa mãn )

\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow a=1,b=2,c=3,d=4\)

Ta có: \(\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{\frac{43}{30}}=\frac{30}{43}\)

Vậy a=1

      b=2

      c=3

      d=4

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

a=1;b=2;c=3;d=4

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

vậy (a;b;c;d)=(1;2;3;4)

\(\frac{30}{43}=\frac{1}{\frac{43}{30}}=\frac{1}{1+\frac{13}{30}}=\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}\)

Vậy a=1 ; b=2 ; c=3 ; d=4

https://olm.vn/images/myolm.png

\(a=1;b=2;c=3;d=4\)

\(\frac{30}{43}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{\frac{43}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow\frac{1}{1+\frac{13}{30}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{\frac{30}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{4}{13}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{\frac{13}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow\frac{1}{1+\frac{1}{2+\frac{1}{3+\frac{1}{4}}}}=\frac{1}{a+\frac{1}{b+\frac{1}{c+\frac{1}{d}}}}\)

\(\Rightarrow a=1,b=2,c=3,d=4\)

đọc mà nổ đầu mất