Xét tử số có dạng : \(\frac{1}{\left(2n+1\right)\left(2n+2\right)\left(2n+3\right)}=\frac{1}{4}\left[\frac{1}{\left(2n+1\right)\left(2n+2\right)}-\frac{1}{\left(2n+2\right)\left(2n+3\right)}\right]\) với \(n\in N\)

Ta có : \(\frac{1}{1.3.5}+\frac{1}{3.5.7}+\frac{1}{5.7.9}+...+\frac{1}{2005.2007.2009}\)

\(=\frac{1}{4}.\left(\frac{1}{1.3}-\frac{1}{3.5}\right)+\frac{1}{4}.\left(\frac{1}{3.5}-\frac{1}{5.7}\right)+\frac{1}{4}\left(\frac{1}{5.7}-\frac{1}{7.9}\right)+...+\frac{1}{4}\left(\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}\left(\frac{1}{1.3}-\frac{1}{3.5}+\frac{1}{3.5}-\frac{1}{5.7}+\frac{1}{5.7}-\frac{1}{7.9}+...+\frac{1}{2005.2007}-\frac{1}{2007.2009}\right)\)

\(=\frac{1}{4}.\left(\frac{1}{3}-\frac{1}{2007.2009}\right)\)

Xét mẫu số có dạng : \(\frac{1}{\left(2n+1\right)\sqrt{2n+3}+\left(2n+3\right)\sqrt{2n+1}}=\frac{1}{\sqrt{2n+1}.\sqrt{2n+3}\left(\sqrt{2n+1}+\sqrt{2n+3}\right)}\)

\(=\frac{\sqrt{2n+3}-\sqrt{2n+1}}{\sqrt{2n+1}.\sqrt{2n+3}\left[\left(2n+3\right)-\left(2n+1\right)\right]}=\frac{1}{2}.\left(\frac{1}{\sqrt{2n+1}}-\frac{1}{\sqrt{2n+3}}\right)\)với  \(n\in N\)

Áp dụng : \(\frac{1}{1\sqrt{3}+3\sqrt{1}}+\frac{1}{3\sqrt{5}+5\sqrt{3}}+\frac{1}{5\sqrt{7}+7\sqrt{5}}+...+\frac{1}{2007\sqrt{2009}+2009\sqrt{2007}}\)

\(=\frac{1}{2}\left(\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{5}}+\frac{1}{\sqrt{5}}-\frac{1}{\sqrt{7}}+...+\frac{1}{\sqrt{2007}}-\frac{1}{\sqrt{2009}}\right)\)

\(=\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)\)

Suy ra : \(M=\frac{\frac{1}{4}\left(\frac{1}{3}-\frac{1}{2007.2009}\right)}{\frac{1}{2}\left(1-\frac{1}{\sqrt{2009}}\right)}\)

Tới đây bài toán đã gọn hơn , bạn tự tính nhé :)

\(S=\left(\frac{-1}{7}\right)^0+\left(\frac{-1}{7}\right)^1+...........+\left(\frac{-1}{7}\right)^{2007}\)

\(\Rightarrow\frac{-1}{7}S=\left(\frac{-1}{7}\right)^1+\left(\frac{-1}{7}\right)^2+........+\left(\frac{-1}{7}\right)^{2008}\)

\(\Rightarrow\frac{-1}{7}S-S=\frac{-8}{7}S=\left(\frac{-1}{7}\right)^{2008}-\left(\frac{-1}{7}\right)^0\)

\(\Rightarrow S=\frac{\left(\frac{-1}{7}\right)^{2008}-1}{\frac{-8}{7}}\)

S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007

=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006

=>(7-1)S=6-(1/7)^2007

=>S=1-(-1/7^2007/6)

1/7S=(-1/7)^1+...+(-1/7)2018

1/7S-S=(-1/7)^1+....+(-1/7)^2018-(-1/7)^0-...-(-1/7)^2017

-6/7S=(-1/7)^2018-1=(-1/7)^2018-1:-6/7

Nguyễn Huy Thắng giải giúp mjnk bài này vs

S=

7S = 1+(−1/7)^1+(−1/7)^2+...+(−1/7)^2007

=> 7S = 7+(−1/7)^1+(−1/7)^2+...+(−1/7)^2006

=> 6S = 6-(−1/7)^2007

=> S= 1-(−1/7^2007/6)

tinh -1/7S rồi lấy -1/7.S-S=-8/7.S=..

\(S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^1+...+\left(-\frac{1}{7}\right)^{2007}\)

\(-\frac{1}{7}S=\left(-\frac{1}{7}\right)^1+\left(-\frac{1}{7}\right)^2+...+\left(-\frac{1}{7}\right)^{2008}\)

\(-\frac{1}{7}S-S=\left(-\frac{1}{7}\right)^0+\left(-\frac{1}{7}\right)^{2008}\)

\(-\frac{8}{7}S=1+\frac{\left(-1\right)^{2008}}{7^{2008}}=1+\frac{1}{7^{2008}}=\frac{7^{2008}+1}{7^{2008}}\)

\(S=\frac{7^{2008}+1}{7^{2008}}:\left(-\frac{8}{7}\right)\)

HOK TOT

a)S=1+(-1/7)^1+(-1/7)^2+...+(-1/7)^2007

=>7S=7+(-1/7)^1+(1/7)^2+...+(-1/7)^2006

=>(7-1)S=6-(1/7)^2007

=>S=1-(-1/7^2007/6)

S=1-1/7-(1/7)^3-......-(1/7)^2017

49S=49-7-1/7-(1/7)^3-.,.....-(1/7)^2015

49S-S=48S=49-7-1-(1/7)^2017

48S=41-(1/7)^2017

S=41/48-(1/7)^2017/48

k nha

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