1^3+2^3+3^3+....+2016^3=S tinh S
1^3+2^3+3^3+2016^3=S tinh S
Tinh S = 3^0+3^1+3^2+..+3^2016
S = 30 + 31 + 32 + ... + 32016
3S = 31 + 32 + 33 + ... + 32016 + 32017
3S - S = ( 31 + 32 + 33 + ... + 32016 + 32017 ) - ( 30 + 31 + 32 + ... + 32016 )
2S = 32017 - 30
2S = 32017 - 1
S = \(\dfrac{3^{2017}-1}{2}\)
tinh tong S = 1-2+3-4+...+2016-2016
Vậy mình nói thảng nhé. Đề này sai số rồi.
Đề đúng như thế này mới hợp lí chứ:
S = 1-2+3-4+...+2015-2016
S= (1-2)+ (3-4) +...+ (2015-2016)
S= (-1)+ (-1)+...+ (-1)
S có 2016 số hạng => Có 1008 số (-1)
=> S= 1008. (-1) =-1008
Vậy S= -1008
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
Tinh:
S=2015 + 2015/1+2 +2015/1+2+3 + 2015/1+2+3+4 +... + 2015/1+2+3+...+2016
Tinh tong S=1/1×2+1/2×3+1/3×4+...+1/2014×2015+1/2015×2016
\(S=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{2015.2016}\)
\(S=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2015}-\frac{1}{2016}\)
\(S=1-\frac{1}{2016}=\frac{2015}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-........+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\left(-\frac{1}{2}+\frac{1}{2}\right)+\left(-\frac{1}{3}+\frac{1}{3}\right)+......+\left(-\frac{1}{2015}+\frac{1}{2015}\right)-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
a^2+b^2+c^2=a^3+b^3+c^3=1. tinh S=a^2+b^2016+c^2017
Tinh tong S=1/1×2+1/2×3+1/3×4+...+1/2014×2015+1/2015×2016
S = 1/1x2 + 1/2x3 + 1/3x4 + ... + 1/2014x2015 + 1/2015x2016
S = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/2014 - 1/2015 + 1/2015 - 1/2016
S = 1 - 1/2016
S = 2015
\(S=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-......+\frac{1}{2015}-\frac{1}{2016}\)
\(S=\frac{1}{1}-\frac{1}{2016}=\frac{2015}{2016}\)
S= 1+5+9+13+........+2013 +2017 tính tổng
cho A= 1+3+3^2+3^3+....+3^2016 và B = 3^2017 tinh B-A
a) Ta có:
S = 1 + 5 + 9 + 13 + ... + 2013 + 2017
S = (2017 + 1)[(2017 - 1) : 4 + 1] : 2
S = 2018.505 : 2
S = 1019090 ÷ 2
S = 509545
b) Ta có:
A = 1 + 3 + 32 + 33 + ... + 32016
3A = 3 + 32 + 33 + 34 + ... + 32017
3A - A = (3 + 32 + 33 + 34 + ... + 32017) - (1 + 3 + 32 + 33 + ... + 32016)
2A = 32017 - 1
A = \(\frac{3^{2017}-1}{2}\)
=> B - A = 32017 - \(\frac{3^{2017}-1}{2}\)
=> B - A = 32017 - \(\frac{3^{2017}}{2}-\frac{1}{2}\)
=> B - A = \(\frac{3^{2017}}{2}-0,5\)