vì sao \(\frac{2}{n\left(n+1\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
Những câu hỏi liên quan
Tại sao lại có công thức này?
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{n+2-n}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)( đpcm )
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\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
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Chứng minh:\(\left(1-\frac{2}{6}\right)\left(1-\frac{2}{12}\right)\left(1-\frac{2}{20}\right)....\left(1-\frac{2}{n\left(n+1\right)}\right)>\frac{1}{3}\)
n thuộc n sao
hi !! ta cũng đang hỏi câu này -_-
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Với n thuộc N* chứng minh
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
CM : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\)
Có : \(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}.\frac{\left(n+2\right)-n}{n\left(n+1\right)\left(n+2\right)}\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}\)\(=\frac{1}{2}\left[\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}\right]\)
\(\frac{1}{n\left(n+1\right)\left(n+2\right)}=\frac{1}{2}\left[\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\right]\) đpcm
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\(\left(\frac{2}{2.3}-1\right)\left(\frac{2}{3.4}-1\right)\left(\frac{2}{4.5}\right)........\left(\frac{2}{n\left(n+1\right)}-1\right)\left(n\in N\ne0,n\ge2\right)\)
2 tháng 4 2018 lúc 21:44
Chứng minh :
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Ta có : \(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{n+2}{n\left(n+1\right)\left(n+2\right)}-\frac{n}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
Vì VT=VP nên ta có đpcm
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\(\text{Ta có:}\)
\(\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}=\frac{1}{n}-\frac{1}{n+1}-\frac{1}{n+1}+\frac{1}{n+2}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(1\right)\)
\(\frac{2}{n\left(n+1\right)\left(n+2\right)}=\frac{2}{n\left(n+1\right)\left(n+2\right)}+\frac{2}{n+1}-\frac{2}{n+1}=\frac{2n\left(n+2\right)+2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)^2}{n\left(n+1\right)\left(n+2\right)}-\frac{2}{n+1}=\frac{2\left(n+1\right)}{n\left(n+2\right)}-\frac{2}{n+1}\left(2\right)\)
\(\text{Từ (1) và (2) ta có: ĐPCM}\)
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help me! (ngu toàn tập)a)frac{3}{left(1.2right)^2}+frac{5}{left(2.3right)^2}+...+frac{2n+1}{left[nleft(n+1right)right]^2}b)left(1-frac{1}{2^2}right)left(1-frac{1}{3^2}right)left(1-frac{1}{4^2}right)....left(1-frac{1}{n^2}right)c)frac{150}{5.8}+frac{150}{8.11}+frac{150}{11.14}+...+frac{150}{47.50}d)frac{1}{1.2.3}+frac{1}{2.3.4}+frac{1}{3.4.5}+...+frac{1}{left(n-1right)nleft(n+1right)}
Đọc tiếp
help me! (ngu toàn tập)
a)\(\frac{3}{\left(1.2\right)^2}+\frac{5}{\left(2.3\right)^2}+...+\frac{2n+1}{\left[n\left(n+1\right)\right]^2}\)
b)\(\left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{4^2}\right)....\left(1-\frac{1}{n^2}\right)\)
c)\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+...+\frac{150}{47.50}\)
d)\(\frac{1}{1.2.3}+\frac{1}{2.3.4}+\frac{1}{3.4.5}+...+\frac{1}{\left(n-1\right)n\left(n+1\right)}\)
\(\frac{150}{5.8}+\frac{150}{8.11}+\frac{150}{11.14}+.....+\frac{150}{47.50}\)
\(=50.\left(\frac{3}{5.8}+\frac{5}{8.11}+.....+\frac{3}{47.50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{11}+......+\frac{1}{47}-\frac{1}{50}\right)\)
\(=50.\left(\frac{1}{5}-\frac{1}{50}\right)\)
\(=50.\frac{9}{50}=9\)
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Cho \(n^4+\frac{1}{4}=\left(\left(n-1\right)n+\frac{1}{2}\right)\left(\left(n+1\right)n+\frac{1}{2}\right)\)
Thu gọn phân thức:
\(\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(13^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(14^4+\frac{1}{4}\right)}\)
Chứng minh :
a) \(\frac{1}{\left(n-1\right)n\left(n+1\right)}=\frac{1}{2}\left(\frac{1}{\left(n-1\right)n}-\frac{1}{n\left(n+1\right)}\right)\)
Chứng minh: \(\frac{3}{\left(1x2\right)}+\frac{5}{\left(2x3\right)}+...+\frac{2n+1}{\left(n\left(n+1\right)\right)^2}=\frac{n\left(n+2\right)}{\left(n+1\right)^2}\)