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Ta có:
\(A=\frac{1}{3}+\frac{2}{3^2}+\frac{3}{3^3}+\frac{4}{3^4}+...+\frac{100}{3^{100}}\)
\(\Rightarrow3A=1+\frac{2}{3}+\frac{3}{3^2}+\frac{4}{3^3}+...+\frac{100}{3^{99}}\)
\(\Rightarrow2A=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}-\frac{100}{3^{100}}\)
\(\Rightarrow6A=3+1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{98}}-\frac{100}{3^{99}}\)
\(\Rightarrow4A=3-\frac{101}{3^{99}}+\frac{100}{3^{100}}=3-\frac{203}{3^{100}}\)
\(\Rightarrow A=\frac{3-\frac{203}{3^{100}}}{4}=\frac{3}{4}-\frac{203}{3^{100}.4}< \frac{3}{4}\Rightarrowđpcm\)
Vậy \(A< \frac{3}{4}\)
\(m=1+3+3^2+...+3^{100}\)
\(3m=3+3^2+...+3^{101}\)
\(\Rightarrow3m-m=3^{101}-1\)
\(\Leftrightarrow m=\frac{3^{101}-1}{2}\)
\(M=1+3+3^2+...+3^{100}\)
\(\Rightarrow\)\(3M=3+3^2+3^3+...+3^{101}\)
\(-M=1+3+3^2+...+3^{100}\)
\(\Rightarrow\)\(2M=3^{101}-1\)
\(\Rightarrow M=\frac{3^{101}-1}{2}\)
\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.6.7}=\frac{\left(-2\right)^2.3.3^2.5^3.7.8}{3.5^3.2\text{^2}.2^2.6.7}=\frac{3^2.8}{2^2.6}=\frac{9.8}{4.6}=\frac{3.3.4.2}{4.3.2}=3\)
Ta có
\(\frac{\left(-2\right)^3.3^3.5^3.7.8}{3.5^3.2^4.42}=\frac{\left(-2\right)^3.3^2.7.8}{2^4.7.6}=\frac{-1.3^2.4}{2.3}=-1.3.2=-6\)
a, 100 + 98 + 96 + ... + 2 - 97 - 95 - 93 - ... - 1
= (100 + 98 + 96 + ... + 2) - (97 + 95 + 93 + ... + 1)
= 2550 - 2401
= 149
b, đặt A = 2 + 22 + 23 + ... + 2100
2A = 22 + 23 + 24 + ... + 2101
2A - A = (22 + 23 + 24 + ... + 2101) - (2 + 22 + 23 + ... + 2100)
A = 2101 - 2
c, 3.32.33....3100
= 31 + 2 + 3 + ... + 100
= 35050
21 + 22 + 23 + ... + 2100
Ta có : S = 2 + 22 + 23 + ... + 2100
2S = 2.(2 + 22 + 23 + ... + 2100)
2S = 22 + 23 + ... + 2100 + 2101
2S - S = (22 + 23 + ... + 2100 + 2101) - (2 + 22 + 23 + ... + 2100)
S = 2101 - 2
\(2^1+2^2+2^3+...+2^{100}\)
Ta có : \(S=2+2^2+2^3+....2^{100}\)
: \(2S=2.\left(2+2^2+2^3+....+2^{100}\right)\)
: \(2S=2^2+2^3+.....+2^{100}+2^{101}\)
: \(2S-S=\left(2^2+2^3+....+2^{100}+2^{101}\right)\)\(-\left(2+2^2+2^3+.....+2^{100}\right)\)
: \(S=2^{101}-2\)
Ta có: \(1^3+2^3+3^3+....+100^3\)
\(=1+2+1.2.3+3+2.3.4+...+100+99.100.101\)
\(=\left(1+2+3+....+100\right)+\left(1.2.3+2.3.4+....+99.100.101\right)\)
\(=5050+101989800\)
\(=101994850\)
bn bảo thế ý là sao