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\(A=2014.\left(1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2013}\right)\)
\(A=2014.\left(1+\frac{1}{3}+\frac{1}{6}+...+\frac{1}{1007.2013}\right)\)
\(A=2.2014.\left(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2013.2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}-\frac{1}{2014}\right)\)
\(A=2.2014.\left(1-\frac{1}{2014}\right)\)
\(A=2.2014.\frac{2013}{2014}\)
\(A=\frac{2.2014.2013}{2014}\)
\(A=2.2013\)
\(A=4026\)
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 \(=\frac{2014^{2016}-1}{2013}\)
\(1.2.3....2015-1.2.3....2014-1.2.3....2013.2014^2\)
\(=1.2.3...\left(2014+1\right)-1.2.3...\left(2014+1\right)\)
\(=0\)
=[1*(1+2)/(1+2)-1/(1+2)]*...*[1*(1+2+3+...+2014)/(1+2+3+...+2014)-1/(1+2+3+...+2014]
còn lại là tịt
Câu hỏi của Phan Nguyễn Hà Linh - Toán lớp 6 - Học toán với OnlineMath
Lời giải:
$M=1+\frac{1}{2}.\frac{2(2+1)}{2}+\frac{1}{3}.\frac{3(3+1)}{2}+\frac{1}{4}.\frac{4(4+1)}{2}+....+\frac{1}{2014}.\frac{2014(2014+1)}{2}$
$=1+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}+...+\frac{2015}{2}$
$=\frac{2+3+4+....+2015}{2}$
$=\frac{1+2+3+....+2015}{2}-\frac{1}{2}$
$=\frac{2015(2015+1)}{4}-\frac{1}{2}=\frac{2031119}{2}$
Đặt S = 1+ (1+2) + (1+2+3) + (1+2+3+4)+......+(1+2+3+....+2014)
S = 1.2 : 2 + 2.3:2 + 3.4:2+......+2014.2015:2
S = (1.2 + 2.3 + 3.4 + ....... + 2014.2015) : 2
Đặt E = 1.2 + 2.3 + 3.4+.....+2014.2015
3E = 1.2.3 + 2.3.(4-1)+...... + 2014.2015.(2016 - 2013)
3E = 1.2.3 + 2.3.4 - 1.2.3 +..... + 2014.2015.2016-2013.2014.2015
3E = (1.2.3 - 1.2.3) +(2.3.4 - 2.3.4)+......+(2013.2014.2015 - 2013.2014.2015) + 2014.2015.2016
3E = 2014.2015.2016
E = 2014*2015*2016/3
< = > S = 2014*2015*2016/3 : 2
S = 1363558560