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\(S=\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}\)
\(S=1-\frac{1}{2!}+\frac{1}{2!}-\frac{1}{3!}+\frac{1}{3!}-\frac{1}{4!}+...+\frac{1}{99!}-\frac{1}{100!}\)
\(S=1-\frac{1}{100!}< 1\)
Vậy S<1
thánh đây rồi , đơn giản vậy em nghĩ mãi k ra , cảm ơn anh nhiều
\(M=5^{2016}-5^{2015}-5^{2014}-...-5-1\)
=>\(5M=5\left(5^{2016}-5^{2015}-5^{2014}-...-5-1\right)\)
=>\(5M=5^{2017}-5^{2016}-5^{2015}-...-5^2-5\)
=>\(5M-M=\left(5^{2017}-5^{2016}-5^{2015}-...-5^2-5\right)-\left(5^{2016}-5^{2015}-5^{2014}-...-5-1\right)\)
=>\(4M=5^{2017}-2.5^{2016}+1\)
=>\(M=\frac{5^{2017}-2.5^{2016}+1}{4}\)
\(5^{2016}-5^{2015}-....-5-1\)
\(=5^{2016}-\left(5^{2015}+5^{2014}+....+5+1\right)\)
\(=5^{2016}-\left(5^{2016}-1\right)\)
\(=5^{2016}-5^{2016}+1\) \(=0+1=1\)
<br class="Apple-interchange-newline"><div id="inner-editor"></div>5M=5(52016−52015−52014−...−5−1)
=>5M=52017−52016−52015−...−52−5
=>5M−M=(52017−52016−52015−...−52−5)−(52016−52015−52014−...−5−1)
=>
Đặt \(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}+\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Nhân \(\frac{1}{7^2}\)vào A. Ta được:
\(A.\frac{1}{7^2}=\frac{1}{7^4}-\frac{1}{7^6}+\frac{1}{7^8}-...-\frac{1}{7^{98}}+\frac{1}{7^{100}}+\frac{1}{7^{102}}\)
\(A=\frac{1}{7^2}-\frac{1}{7^4}+\frac{1}{7^6}-\frac{1}{7^8}+...+\frac{1}{7^{98}}-\frac{1}{7^{100}}\)
Ta có: \(\frac{1}{7^2}.A+A=\frac{1}{49}-\frac{1}{7^{102}}\Rightarrow\frac{50}{49}.A=\frac{1}{49}-\frac{1}{7^{102}}\)
\(\Rightarrow A=\left(\frac{1}{49}-\frac{1}{7^{102}}\right)\frac{49}{50}< \frac{1}{5}^{\left(đpcm\right)}\)
\(A=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{16}\)
\(A=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)+\left(\frac{1}{9}+\frac{1}{10}+\frac{1}{11}+\frac{1}{12}\right)+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}\right)\)
\(>1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+4\times\frac{1}{8}+4\times\frac{1}{12}+4\times\frac{1}{16}\)
\(=1+\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\)
\(=1+2\times\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}\right)\)
\(>1+2\times\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{4}\right)=1+2\times1\)
\(=1+2=3=B\)
\(\Rightarrow A>B\)
Học tốt
ta có : 87-218=221-218=218(23-1)=218-7chia hết cho 2 và 7 . vậy nó chia hết cho 14
vì 2-1=1
1+1=2 vì 2-1=1