1/3-2/3^2+3/3^3-4/3^4.......2014/3^2014 >1/5
CMR : 1/3-2/3^2+3/3^3-4/3^4+...-2014/3^2014<1/5
2014+(2014/1+2)+(2014/1+2+3)+...+(2014/1+2+3+4+5+...+2013)=???.Ai giai duoc vay???
2014+(2014/1+2)+(2014/1+2+3)+...+(2014/1+2+3+...+2013)
=2014*(1+(1/1+2)+(1/1+2+3)+...+( 1/1+2+3+...+2013))
=2014*(1+(1/3)+(1/6)+....+(1/2027091)
=2014*2*((1/+(1/2*3)+(1/3*4).....+(1/2013*2014))
=2014*2*(1/1-1/2+1/2-1/3+1/3-1/4+.....+1/2013-1/2014)
=2014*2*(1-1/2014)
=2*(2014*2013/2014)
=2*2013
=4026
Cuối cùng cũng giải được.
Bài 1 : Tính tổng
a) 1 *2 *3 + 2 * 3 *4 + 3 * 4 * 5 + ... + 2013 * 2014 * 2015 + 2014 * 2015 * 2016
b) 1 * + 3 * 4 + 5 * 6 + ... + 99 * 100
Bài 2 : CMR : 1^3 + 2^3 + 3^3 + ... + n^3 = ( 1 + 2 + 3 + ... + n )^2
A=2014+[2014:(1+2)]+[2014:(1+2+3)]+[2014:(1+2+3+4)]+...++[2014:(1+2+3+...+2013)]
(1/2×3+1/3×4+1/4×5) :19/20×2014×+1=2014
bài này hơi khó,mình ko giai được,xin lỗi bạn nhé
Tính tổng:
A= 1+2014^1+2014^2+2014^3+...+2014^2014+2014^2015
B = 3-3^2+3^3+3^4+...+3^100
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 \(=\frac{2014^{2016}-1}{2013}\)
B = 3 - 32 + 33 - 34 + ... + 3100
3B= 32 - 33 + 34 - 35 + ... + 3101
3B + B = ( 3 - 32 + 33 - 34 + ... + 3100 ) + ( 32 - 33 + 34 - 35 + ... + 3101 )
4B = 3 + 3101
B = \(\frac{3+3^{101}}{4}\)
Rút gọn biểu thức : \(\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+2}+\frac{3}{3^4+3^2+3}+...+\frac{2014}{2014^4+2014^2+2014}\)
cm 1/3-2/3^2+3/3^3-4/3^4+.....-2014/3^2014<1/3
CMR:\(\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}+...-\frac{2014}{3^{2014}}<\frac{1}{5}\)