SS : \(2^0\)+\(2^1\)+\(2^2\) + .....+\(2^{100}\) voi \(2^{101}\)
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SS : \(^{2^0}\) + \(^{2^1}\) + \(^{2^2}\) +...+ \(^{2^{100}}\) voi \(2^{101}\)
Tinh tong sau:
N=1^1+2^2+3^3+...+100^100. So sanh N voi 101^102.
\(A=2^0+2^1+2^2+2^3+2^4+...+2^{100}\)
\(B=2^{101}\)
\(Ta\) \(có:\)
\(A=2^0+2^1+2^2+2^3+...+2^{100}\)
\(=>2A=2+2^2+2^3+...+2^{101}\)
\(=>2A-A=\left(2+2^2+2^3+...+2^{101}\right)-\left(2^0+2^1+2^2+...+2^{100}\right)\)
\(=>A=2^{101}-1\)
\(Mà\) \(B=2^{101}\)
\(=>A< B\)
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A=1*2*3+3*4*5+...+99*100*101 ; B=1*2²+2*3²+3*4²+...+99*100²
Giúp mau cho e voi, em dang rat can , dup e di
20+21+....+2100........2101
Đặt :
\(A=2^0+2^1+....+2^{101}\)
\(\Rightarrow2A=2+2^2+...+2^{102}\)
\(\Rightarrow2A-A=\left(2+2^2+...+2^{102}\right)-\left(2^0+2^1+....+2^{101}\right)\)
\(\Rightarrow A=2^{102}-1\)
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Gọi biểu thức trên là A
Ta có:
A = 20 + 21 + .... + 2100 + 2101
\(\Rightarrow\) 2A = 2 . (20 + 21 + .... + 2100 + 2101)
\(\Rightarrow\) 2A = 2 + 22 + .... + 2101 + 2102
\(\Rightarrow\) 2A - A = (2 + 22 + .... + 2101 + 2102) - (20 + 21 + .... + 2100 + 2101)
\(\Rightarrow\) A = 2102 - 1
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A1+frac{3}{2^3}+frac{4}{2^4}+...+frac{100}{2^{100}}frac{A}{2}frac{1}{2}+frac{3}{2^4}+frac{4}{2^5}+....+frac{100}{2^{101}}A-frac{A}{2}left(1+frac{3}{2^3}+....+frac{100}{2^{100}}right)-left(frac{1}{2}+frac{3}{2^4}+.....+frac{100}{2^{101}}right)frac{A}{2}frac{1}{2}+frac{3}{2^3}+frac{1}{2^4}+frac{1}{2^5}+....+frac{1}{2^{100}}-frac{100}{2^{101}}frac{A}{2}frac{1}{2}+frac{1}{2^2}+frac{1}{2^3}+frac{1}{2^4}+....+frac{1}{2^{100}}-frac{1}{2^{101}}frac{A}{2}left(1-left(frac{1}{2}right)^{101}right).2-frac{10...
Đọc tiếp
\(A=1+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^4}+\frac{4}{2^5}+....+\frac{100}{2^{101}}\)\(A-\frac{A}{2}=\left(1+\frac{3}{2^3}+....+\frac{100}{2^{100}}\right)-\left(\frac{1}{2}+\frac{3}{2^4}+.....+\frac{100}{2^{101}}\right)\)
\(\frac{A}{2}=\frac{1}{2}+\frac{3}{2^3}+\frac{1}{2^4}+\frac{1}{2^5}+....+\frac{1}{2^{100}}-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}-\frac{1}{2^{101}}\)
\(\frac{A}{2}=\left(1-\left(\frac{1}{2}\right)^{101}\right).2-\frac{100}{2^{101}}\)
\(\frac{A}{2}=\frac{2^{101}-1}{2^{100}}-\frac{100}{2^{101}}\)
\(A=\frac{2^{101}-1}{2^{99}}-\frac{100}{2^{100}}\)
1, cho a^100+b^100=a^101+b^101=a^101+b^101=a^102+b^102.CM a+b/b=a^2+b^2/a^2b^2
2,tính gtbt:A= x/xy+x+1+y/y+1+yz+z/1+z+xz
3, cho a,b,c,d>0 TM:a^2+b^2=1 và a^4/b+c^4/d=1/b+d CM:a^2016/b^1003+c^2006/d^1003=2/(b+d)^1003
so sánh A và B biết A=2^0 + 2^1 + 2^2 +...+ 2^100 B=2^101
Cho ham so
f(x)=4^x/4^x+2
Tinh A=f(0)+f(1/101)+f(2/101)+f(3/101)+...+f(100/101)+f(1)
A=f(0)+(f(1/101)+f(100/101))+(f(2/101)+f(99/101))+...+f(1)
A=f(0)+50f(1)+f(1)
A=f(0)+51f(1)
A=4^0/4^0+2+51(4^1/4^1+2)
A=1/3+34
A=103/3
Mik ko bik đúng ko nữa
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