Chứng minh rằng:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{98.98.100}=\frac{4949}{19800}\)
CHỨNG TỎ: \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{4949}{19800}\)
Lưu ý: ko pk toán lớp1
\(\frac{2\left(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}\right)}{2}\)
(\(\frac{2}{1\cdot2\cdot3}+\frac{2}{2\cdot3\cdot4}+\frac{2}{3\cdot4\cdot5}+...+\frac{2}{98\cdot99\cdot100}\)) : 2
(\(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot4}+...+\frac{1}{98\cdot99}-\frac{1}{99\cdot100}\)) : 2
mình làm tiếp nha lúc nãy lỡ tay
\(\frac{\left(\frac{1}{2}-\frac{1}{9900}\right)}{2}=\frac{4949}{19800}=VP\)
Vậy ....
Tìm x , biết:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+......+\frac{1}{x\left(x+1\right)\left(x+2\right)}=\frac{4949}{19800}\)
chứng tỏ : \(\frac{1}{1.2.3}\)+\(\frac{1}{2.3.4}\)+\(\frac{1}{3.4.5}\)+.........+\(\frac{1}{98.99.100}\)= \(\frac{4949}{19800}\)
giải gấp mk nha! nếu đúng mh tick cho.
chứng tỏ
1/1.2.3 + 1/2.3.4 + 1/3.4.5 + ... + 1/98.99.100 = 4949/19800
= 1/2.(1/1.2 - 1/2.3 + 1/2.3 - 1/3.4 + 1/3.4 - 1/4.5 + 1/4.5 - ........+1/98.99 - 1/99.100 )
=1/2.(1/1.2 - 1/99.100)
=1/2 . 4949/9900
=4949/19800
Chứng tỏ \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+\frac{1}{3\cdot4\cdot5}+...+\frac{1}{98\cdot99\cdot100}=\frac{4949}{19800}\)
Ta có 1/1.2-1/2.3=2/1.2.3;1/2.3-1/3.4=2/2.3.4 .....1/98.99-1/99.100=2/98.99.100 2A=2/1.2.3+2/2.3.4+....+2/98.99.100 = 1/1.2-1/2.3+1/2.3-1/3.4+...+1/98.99-1/99.100 = 1/2-1/99.100 = 4949/9900 A =4949/19800
chưng tỏ 1/1.2.3 +1/2.3.4 + 1/3.4.5+...+1/98.99.100=4949/19800
Bạn cho sai đề rồi !
Sửa : Chứng tỏ : \(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}=\frac{4949}{9900}\)
Ta có : \(VT=\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{98.99.100}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{98.99}-\frac{1}{99.100}\)
\(=\frac{1}{1.2}-\frac{1}{99.100}\)
\(=\frac{99.100-2}{2.99.100}\)
\(=\frac{4949}{9900}=VP\)
Study well ! >_<
CMR: 1/(1.2.3)+1/(2.3.4)+...+1/(98.99.100)=4949/19800
Cho \(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}.\)
Chứng minh rằng:\(B< \frac{1}{4}.\)
2B=\(\frac{2}{1.2.3}\)+.....+\(\frac{2}{18.19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{2.3}\)+\(\frac{1}{2.3}\)-\(\frac{1}{3.4}\).......+\(\frac{1}{18.19}\)-\(\frac{1}{19.20}\)
2B=\(\frac{1}{1.2}\)-\(\frac{1}{19.20}\)
B=\(\frac{1}{1.2}\):2-\(\frac{1}{19.20}\):2
B=\(\frac{1}{1.2}\).\(\frac{1}{2}\)-\(\frac{1}{19.20}\).\(\frac{1}{2}\)
=\(\frac{1}{4}\)-\(\frac{1}{19.20.2}\)<\(\frac{1}{4}\)
\(B=\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{18.19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{18.19}-\frac{1}{19.20}\)
\(2B=\frac{1}{1.2}-\frac{1}{19.20}\)
\(B=\left(\frac{1}{2}-\frac{1}{19.20}\right):2\)
\(B=\frac{189}{760}\)
Tớ đồng ý với kết quả của Jenny Dolly Marion _ Love For You và Đầu Vụ Công
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tk tớ nhé ( chúc bn học giỏi )
Chứng minh rằng với mọi số nguyên dương n, ta có:
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+....+\frac{1}{n\left(n+1\right)\left(n+2\right)}< \frac{1}{4}\)
\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+...+\frac{1}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}\left(\frac{1}{1.2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right)\)
\(=\frac{1}{4}-\frac{1}{2\left(n+1\right)\left(n+2\right)}\) \(< \frac{1}{4}\)