2015 + (2015/ 1+2) + (2015/ 1+2+3) +......+ (2015/ 1+2+3+...+2014) =?
Tính: (1*2015+2*2014+3*2013+...+2015*1)/(1*2+2*3+3*4+4*5+...+2015*2016)
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016)
Cho A = \(\dfrac{2015}{2014^2+1}+\dfrac{2015}{2014^2+2}+\dfrac{2015}{2014^3+3}+....+\dfrac{2015}{2014^2+2014}\)
Chứng minh rằng A không là số nguyên dương
Các bạn ơi , giúp mình với T T
RGBT:
E=\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2015\sqrt{2014}+2014\sqrt{2015}}+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
Ta có:
\(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n\left(n+1\right)}\left(\sqrt{n+1}+\sqrt{n}\right)}\)
\(=\frac{\left(\sqrt{n+1}-\sqrt{n}\right)}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
Thế vô bài toán được
\(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+...+\frac{1}{2016\sqrt{2015}+2015\sqrt{2016}}\)
\(=\frac{1}{\sqrt{1}}-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}\)
\(=1-\frac{1}{\sqrt{2016}}\)
tìm s
Tính S = 1/2(1+2) + 1/3(1+2+3)+...+ 1/2015(1+2+...+2014+2015) + 1/2016(1+2+...+2015+2016
Cho A = 2015/20142+1 + 2015/20142+2 + 2015/20142+3 +.......+ 2015/20142+2014
Chứng minh rằng Akhông phải là số nguyên dương.
(1/2012+1/2013-1/2014)/(5/2012+5/2013-5/2014)-(2/2103+2/2014-2/2015)/(3/2013+3/2014-3/2015)
\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{\frac{5}{2012}+\frac{5}{2013}-\frac{5}{2014}}-\frac{\frac{2}{2013}+\frac{2}{2014}-\frac{2}{2015}}{\frac{3}{2013}+\frac{3}{2014}-\frac{3}{2015}}\)
=\(\frac{\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}}{5\left(\frac{1}{2012}+\frac{1}{2013}-\frac{1}{2014}\right)}-\frac{2\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}{3\left(\frac{1}{2013}+\frac{1}{2014}-\frac{1}{2015}\right)}=\frac{1}{5}-\frac{2}{3}=\frac{3}{15}-\frac{10}{15}=-\frac{7}{15}\)
Tính các tổng sau:
a) A=1+(-2) + 3 +(-4) + ...+(- 2014) + 2015;
b) B= (-2) + 4 +(-6) + 8 ... +(-2014) + 2016;
c) 1+(-3) + 5 +(-7) + ... + 2013 +(-2015);
d) (-2015) + (-2014) + (-2013)+ ... + 2015 + 2016
\(A=\left[1+\left(-2\right)\right]+\left[3+\left(-4\right)\right]+....+\left[2013+\left(-2014\right)+2015\right]\)
\(A=\left(-1\right)+\left(-1\right)+....+\left(-1\right)+2015\left(\text{1007 số hạng }\left(-1\right)\right)=1008\)
\(B=\left(-2\right)+4+\left(-6\right)+8+\left(-10\right)+,...+\left(-2014\right)+2016\)
\(B=2+2+....+2\left(\text{504 số hạng 2}\right)=1008\)
c) 1 + ( -3 ) +5 + ( -7 ) + ...........+ 2013 + ( -2015 )
[ 1 + (-3 ) ] + [ 5 + -7 ] + .......... + [ 2013 + ( - 2015 ) ]
có số cặp là : [ ( 2015 - 1 ) : 2 + 1 ] : 2 = 504 ( cặp )
= -2 + -2 + -2 +..........+ -2
= -2 x 504
= -1008
A=1/2015+2/2015+3/2015+........+2014/2015
Vậy A bằng.............