2^0+2^1+2^3+...+2014^2014
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\(S=2014+\frac{2014}{1+2}+\frac{2014}{1+2+3}+...+\frac{2014}{1+2+3+...+10000}\)
\(S=\frac{2014}{\frac{1.2}{2}}+\frac{2014}{\frac{2.3}{2}}+\frac{2014}{\frac{3.4}{2}}+...+\frac{2014}{\frac{10000.10001}{2}}\)
\(S=\frac{4028}{1.2}+\frac{4028}{2.3}+\frac{4028}{3.4}+...+\frac{4028}{10000.10001}\)
\(S=4028\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10000.10001}\right)\)
\(S=4028\left(\frac{2-1}{1.2}+\frac{3-2}{2.3}+\frac{4-3}{3.4}+...+\frac{10001-10000}{10000.10001}\right)\)
\(S=4028\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{10000}-\frac{1}{10001}\right)\)
\(S=4028\left(1-\frac{1}{10001}\right)=\frac{40280000}{10001}\)
A = 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015
2014A = 2014^1 + 2014^2 + 2014^3 + 2014^4 + ... 2014^2015 + 2014^2016
2014A - A = ( 2014^1 + 2014^2 + 2014^3 + 2014^4 + .... + 2014^2015 + 2014^2016 ) - ( 1 + 2014^1 + 2014^2 + 2014^3 + ... + 2014^2014 + 2014^2015 )
2013A = 2014^2016 - 1
A = 2014^2016 - 1 / 2013
B = 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 ( đề hơi vui )
3B = 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101
3B - B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - ( 3 - 3^2 + 3^3 + 3^4 + ... + 3^100 )
2B = ( 3^2 - 3^3 + 3^4 + 3^5 + ... + 3^101 ) - 3 + 3^2 - 3^3 - 3^4 - ... - 3^100
2B = 3^2 - 3^3 + 3^101 - 3 + 3^2 - 3^3
2B = 9 - 27 + 3^101 - 3 + 9 - 27
2B = -18 + 3^101 - 3 + ( -18 )
2B = -39 + 3^101
B = -39 + 3^101 / 2
A = 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015
2014A = 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016
2014A - A = ( 2014 + 20142 + 20143 + 20144 + ... + 20142015 + 20142016 ) - ( 1 + 2014 + 20142 + 20143 + ... + 20142014 + 20142015 )
2013A = 20142016 - 1
A \(=\frac{2014^{2016}-1}{2013}\)
\(S=\left(-2\right)^0+\left(-2\right)^1+\left(-2\right)^2+\left(-2\right)^3+....+\left(-2\right)^{2014}+\left(-2\right)^{2015}\)
\(\left(-2\right)S=\left(-2\right)+\left(-2\right)^2+\left(-2\right)^3+\left(-2\right)^4+....+\left(-2\right)^{2016}\)
\(\left(-2\right)S-S=\left[\left(-2\right)+\left(-2\right)^2+...+\left(-2\right)^{2016}\right]-\left[1+\left(-2\right)^1+...+\left(-2\right)^{2015}\right]\)
\(S=\left(-2\right)^{2016}-1\)
\(2^0+2^1+2^2+2^3+...+2^{2014}.\)
\(=1+\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+.....+\left(2^{2012}+2^{2013}+2^{2014}\right)\)
\(=1+2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{2012}\left(1+2+2^2\right)\)
\(=1+2.7+2^4.7+.....+2^{2012}.7\)
\(=1+7\left(2+2^4+....+2^{2012}\right)\)
\(7\left(2+2^4+...+2^{2012}\right)⋮7\)\(\Rightarrow\)\(2^0+2^1+2^2+2^3+...+2^{2014}\)\(chia7\)\(dư1\)
A = 2014 (\(1+\frac{1}{1+2}+\frac{1}{1+2+3}+.....+\frac{1}{1+2+3+....+2013}\))
A = 2014(1+1/3 + 1/6 +....+ 1/1007.2013)
A = 2014( 2/2 + 2/6 + 2/12 +.....+ 2/2013.2014)
A = 2.2014( 1/2 + 1/6 +....+ 1/2013.2014)
A = 2.2014( 1/1.2 + 1/2.3 +.....+ 1/2013.2014)
A = 2.2014( 1 - 1/2 + 1/2 - 1/3 +.....+ 1/2013 - 1/2014)
A = 2.2014( 1 - 1/2014)
A = 2.2014 . 2013/2014
A = 2.2014.2013/2014
A = 4026
Câu hỏi của h - Chuyên mục hỏi đáp - Giúp tôi giải toán. - Học toán với OnlineMath
Ta có: \(S=\left(-\dfrac{1}{7}\right)^0+\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+...+\left(-\dfrac{1}{7}\right)^{2014}\)
\(\Leftrightarrow\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^1+\left(-\dfrac{1}{7}\right)^2+\left(-\dfrac{1}{7}\right)^3+...+\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow S-\dfrac{-1}{7}\cdot S=\left(-\dfrac{1}{7}\right)^0-\left(-\dfrac{1}{7}\right)^{2015}\)
\(\Leftrightarrow\dfrac{8}{7}\cdot S=1+\dfrac{1}{7^{2015}}\)
\(\Leftrightarrow S=\left(1+\dfrac{1}{7^{2015}}\right):\dfrac{8}{7}=\dfrac{\left(1+\dfrac{1}{7^{2015}}\right)\cdot7}{8}\)