a, S= 3+\(^{ }3^2\)+\(3^3\)+....+\(^{ }3^{100}\)
3xS= \(3^2+3^3+...+3^{100}+3^{101}\)
3xS - S= \(3^{101}\)-3
2xS= \(3^{101}\)-3
S= \(3^{101}\)-3/2
Ta xét:
\(3^{101}\)= \(\left(3^4\right)^{25}\)x3= \(81^{2005}\) x3=(...1) x (...3)=(...3)
Vậy chữ số tận cùng của S là 1.
Chúc bạn học có hiệu quả!
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a) CM: A2= \(\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{100}}>10\)
b) CM: A3= \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+\frac{4}{5!}+...+\frac{99}{100!}< 1\)
CMR:
a) \(\frac{1}{2!}+\frac{2}{3!}+\frac{3}{4!}+...+\frac{99}{100!}< 1\)
b) \(\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+\frac{3.4-1}{4!}+...+\frac{99.100-1}{100!}< 2\)
Chứng minh: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...................+\frac{100}{3^{100}}< \frac{3}{4}\)
Chứng minh: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...................+\frac{100}{3^{100}}< \frac{3}{4}\)
Chứng minh rằng: \(\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+....................+\frac{100}{3^{100}}< \frac{3}{4}\)
cho biểu thức: \(A=\left(\frac{-1}{3}\right)+\frac{1}{3^2}-\frac{1}{3^3}+...+\frac{1}{3^{100}}\)
tính \(B=4\left|A\right|+\frac{1}{3^{100}}\)
tính giá trị biểu thức
A=\(\frac{\left[1+2+3+......+100\right].\left[\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-\frac{1}{5}\right].\left[2,4.42-21.4,8\right]}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+....+\frac{1}{100}}\)
tính
a)\(\left(1-\frac{1}{1+2}\right).\left(1-\frac{1}{1+2+3}\right).\left(1-\frac{1}{1+2+3+4}\right).....\left(1-\frac{1}{1+2+3+...+20}\right)\)
b)\(\frac{\left(1+2+3+...+100\right).\left(12.3,4-6,86\right)}{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}}\)
c)(18.123+9.436.2+3.5310.6):1+4+7+...+100-410)
giúp mk vs mai mk kiểm tra rồi . ai đúng mk tick nha
1/ Chứng minh: \(C=\frac{1}{3}-\frac{2}{3^2}+\frac{3}{3^3}-\frac{4}{3^4}....+\frac{99}{3^{99}}-\frac{100}{3^{100}}\). Chứng minh: C < \(\frac{3}{16}\)
