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a)A=\(\frac{\left(8+100\right).\left[\left(100-8\right):4+1\right]}{2}=\frac{108.242}{2}=13068\)
b) \(5B=5^2+5^3+...+5^{101}\)
\(5B-B=5^{101}-5\)
\(B=\frac{5^{101}-5}{4}\)
A=(1-1/2^2)(1-1/3^2).....(1-1/10^2)
=(1-1/4)(1-1/9)....(1-1/100)
=3/4.8/9......99/100=1.3.2.4.....9.11/(2.3.....10)^2=1.2.3.....11/(3.....10).10=2.11/10=22/10=11/5
B=1/1.3+1/3.5+1/5.7+1/7.9
= 2.(1/1.3+1/3.5+1/5.7+1/7.9).1/2=(1/1-1/3+1/3-1/5+1/5-1/7+1/7-1/9).1/2=(1/1-1/9).1/2=8/9.1/2=4/9
a/M=2/3.5+2/5.7+2/7.9+.....+2/97.99
M=1/3-1/5+1/5-1/7+..+1/97-1/99
M=1/3-1/99
M=32/99
b)ta có 1/2.3+1/3.4+1/4.5+..+1/2015.2016+1/2016.2017<A
=>1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016+1/2016-1/2017<a
1/2-1/2017<A
2/15/4034<A (1)
Ta có
1/1.2+1/2.3+1/3.4+1/4.5+..+1/2015.2016>A
=>1-1/2+1/2-1/3+1/3-1/4+1/4-1/5+..+1/2015-1/2016>A
1-1/2016
2015/2016>A (2)
Từ (1) và (2)=>A không phải là số tự nhiên(đpcm)
a) \(\left(1^2+2^2+3^2+....+2012^2\right).\left(91-273:3\right)\)
\(=\left(1^2+2^2+3^2+...+2012^2\right).\left(91-91\right)\)
\(=0\)
b) \(\left(-284\right).172+\left(-284\right).\left(-72\right)=\left(-284\right).\left(172+-72\right)\)
\(=\left(-284\right).100\)
\(=-28400\)
c) \(\frac{1}{5}+\frac{-1}{6}+\frac{1}{7}+\frac{-1}{8}+\frac{1}{9}+\frac{1}{8}+\frac{-1}{7}+\frac{1}{6}+\frac{-1}{5}\)
\(=\left(\frac{1}{5}+\frac{-1}{5}\right)+\left(\frac{1}{6}+\frac{-1}{6}\right)+\left(\frac{1}{7}+\frac{-1}{7}\right)+\left(\frac{1}{8}+\frac{-1}{8}\right)+\frac{1}{9}\)
\(=0+0+0+0+\frac{1}{19}\)
= 0
Ta có: \(B=\frac{1}{3\cdot4}+\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}\)
\(=\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}\)
\(=\frac{1}{3}-\frac{1}{8}\)
\(=\frac{5}{24}\)
Vậy \(B=\frac{5}{24}\)
Ta có: \(C=\frac{1}{4\cdot5}+\frac{1}{5\cdot6}+\frac{1}{6\cdot7}+\frac{1}{7\cdot8}+\frac{1}{8\cdot9}\)
\(=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+\frac{1}{7}-\frac{1}{8}+\frac{1}{8}-\frac{1}{9}\)
\(=\frac{1}{4}-\frac{1}{9}\)
\(=\frac{5}{36}\)
Vậy \(C=\frac{5}{36}\)
Bạn làm được mỗi câu 1 à, thôi ko sao