Cho B = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...........+\left(1+2+3+......+98\right)}{1\times98+2\times97+3\times96+.....+98\times1}\)
Rút gọn B
a,A=\(\frac{1}{2}-\frac{1}{2^2}+\frac{1}{2^3}-\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{1}{2^{100}}\)
b,B=\(\frac{1}{1\times2\times3}+\frac{1}{2\times3\times4}+\frac{1}{3\times4\times5}+...+\frac{1}{998\times999\times100}\)
c,C=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\times98+2\times97+3\times96+...+98\times1}\)
Tính:
A = \(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99\right)}{1\times99+2\times98+3\times97+...+99\times1}\)
B = \(\frac{1\times2010+2\times2009+3\times2008+...+2010\times1}{\left(1+2+3+...+2010\right)+\left(1+2+3+...+2009\right)+...+\left(1+2\right)+1}\)
\(\frac{1+< 1+2>+< 1+2+3>+..............+< 1+2+3+4+...........+98>}{1\times98+2\times97+3\times96+..............................+98\times1}\)
Giúp mình nhé ai nhanh nhất mình tíck
Cho B=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1\cdot98+2\cdot97+3\cdot96+...+98\cdot1}\)
Rút gọn B ta được B=
rút gọn \(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+.........+\left(1+2+3+.......+98\right)}{1\cdot98+2\cdot97+3\cdot96+........+98\cdot1}\)
Cho B=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+...+98.1}\). Rút gọn B ta được B=..........
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+...+98.1}\)
\(B=\frac{1+1+2+1+2+3+...+1+2+3+...+98}{1.98+2.97+3.96+...+98.1}\)
\(B=\frac{1.98+2.97+3.96+...+98.1}{1.98+2.97+3.96+...+98.1}\)
\(B=1\)
Cho B=\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+...+98.1}\) Rút gọn B ta được B=
Tính
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+99+100\right)}{\left(1\times100+2\times99+3\times98+...+99\times2+100\times1\right)\times2013}\)
\(\frac{1+\left(1+2\right)+\left(1+2+3\right)+.....+\left(1+2+3+4+......+100\right)}{\left(1.100+2.99+3.98+.......+99.2+100.1\right).2013}\)
\(=\frac{1.100+2.99+3.98+......+99.2+100.1}{\left(1.100+2.99+3.98+.....+99.2+100.1\right).2013}\)
\(=\frac{1}{2013}\)
Cho \(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+...+98.1}\)
Rút gọn B.
Ta có : Có 98 cs 1; 97 cs 2; 96 cs 3;...;1 cs 98 \(\Leftrightarrow\)98.1+97.2+96.3+...+1.98
\(\Rightarrow\)Tử=mẫu
Vậy B= 1 đó bn
\(B=\frac{1+\left(1+2\right)+\left(1+2+3\right)+...+\left(1+2+3+...+98\right)}{1.98+2.97+3.96+..+98.1}\)
Do tử số gồm 98 tổng: số 1 xuất hiện 98 lần; số 2 xuất hiện 97 lân; soos xuất hiện 96 lần;...; số 98 xuất hiện 1 lân nên Ta có : 1+(1+2)+(1+2+3)+...+(1+2+3+...+98)
= 1+1+2+1+2+3+...+1+2+3+...+98
= 1.98+2.97+3.96+...+98.1
=> B = \(B=\frac{1.98+2.97+3.96+...+98.1}{1.98+2.97+3.96+...+98.1}=1\)
Vậy B=1