So sánh (1/2)^4 và (1/4)^4
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4 mũ 15+1/4 mũ 17 +1= 1/16+1
4 mũ 12+1/ 4 mũ 14+1= 1/16+1
suy ra 1/17=1/17
suy ra A=B
nhớ tích cho tớ nhé
\(S=1+4^2+4^3+...+4^{99}\)
\(\Rightarrow S+4=1+4+4^2+4^3+...+4^{99}\)
\(\Rightarrow S+4=\dfrac{4^{99+1}-1}{4-1}=\dfrac{4^{100}-1}{3}\)
\(\Rightarrow S=\dfrac{4^{100}-1}{3}-4=\dfrac{4^{100}-13}{3}\)
\(\Rightarrow3S+1=3.\dfrac{4^{100}-13}{3}+1\)
\(\Rightarrow3S+1=4^{100}-12\)
\(\Rightarrow3S+1=2^{200}-2^2.3>2^{100}\)
mà \(32^{20}=\left(2^5\right)^{20}=2^{100}\)
\(\Rightarrow3S+1>32^{20}\)
A = 1/42 + 1/62 + 1/82 + ... + 1/(2n)2
A = 1/22.(1/22 + 1/32 + 1/42 + ... + n2)
A < 1/22.(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/(n-1).n
A < 1/4.(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... +1/n-1 - 1/n)
A < 1/4.(1 - 1/n) < 1/4.1
A < 1/4
\(32^{15}=\left(2^5\right)^{15}=2^{5.15}=2^{75}\)
\(4^{39}=\left(2^2\right)^{39}=2^{2.39}=2^{78}\)
Do \(2^{78}>2^{75}\)
\(\Rightarrow4^{39}>32^{15}\)
\(\Rightarrow1+4+4^2+...+4^{39}>32^{15}\)
\(\Rightarrow3\left(1+4+4^2+...+4^{39}\right)>32^{15}\)
Vậy \(A>B\)
\(\left(1+2+3+4\right)^2=\left(1+2+3+4\right)\left(1+2+3+4\right)=1^2+1\cdot2+1\cdot3+1\cdot4+2\cdot1+2^2+2\cdot3+2\cdot4+3\cdot1+3\cdot2+3^2+3\cdot4+4\cdot1+4\cdot2+4\cdot3+4^2=1^2+2^2+3^2+4^2+1\cdot2+1\cdot3+1\cdot4+2\cdot1+2\cdot3+2\cdot4+3\cdot1+3\cdot2+3\cdot4+4\cdot1+4\cdot2+4\cdot3>1^2+2^2+3^2+4^2\)
Bài giải
Ta có : \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{2^4}\)
\(\left(\frac{1}{4}\right)^4=\frac{1^4}{4^4}=\frac{1}{4^4}\)
Vì \(2^4< 4^4\text{ }\Rightarrow\text{ }\frac{1}{2^4}>\frac{1}{4^4}\)
Bài giải
Ta có : \(\left(\frac{1}{2}\right)^4=\frac{1^4}{2^4}=\frac{1}{2^4}\)
\(\left(\frac{1}{4}\right)^4=\frac{1^4}{4^4}=\frac{1}{4^4}\)
Vì \(2^4< 4^4\text{ }\Rightarrow\text{ }\frac{1}{2^4}>\frac{1}{4^4}\)