Rút gọn: b) (3.2 ^ (n + 2) + 2 ^ n - 2 ^ (n - 1)) : 5
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VT
0
2 tháng 8 2018
\(a,3^{n+2}-3^{n+1}+6.3^n\)
\(=3^n\left(3^2-3+6\right)=3^n.12\)
\(b,\left(3.2^{n+2}+2^n+2^{n+1}\right):5\)
\(=\left[2^n\left(3.2^2+1+2\right)\right]:5\)
\(=2^n.15:5\)
\(=2^n.3\)
TT
2
A
8 tháng 3 2019
\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-1+\frac{1}{100}\)
\(A=\frac{2}{100}-1\)
\(A=\frac{1}{50}-1\)
\(A=\frac{-49}{50}\)
A
8 tháng 3 2019
b,\(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n=2^{n+34}\) (1)
Đặt \(B=2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\)
\(\Rightarrow2B=2.\left(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n\right)\)
\(=2.2^3+3.2^4+4.2^5+...+\left(n-1\right).2^n+n.2^{n+1}\)
\(2B-B=\left(2.2^3+3.2^4+4.2^5+..+\left(n-1\right).2^n+n.2^{n+1}\right)\)
\(=(2.2^2+3.2^3+4.2^4+...+\left(n-1\right).2^{n-1}+n.2^n)\)
\(B=-2^3-2^4-2^5-...-2^{n+1}-2.2^2\)
\(=-\left(2^3+2^4+2^5+...+2^n\right)+n.2^{n+1}-2^3\)
Đặt \(C=2^3+2^4+2^5+2^n\)
\(\Rightarrow2C=2.(2^3+2^4+2^5+...+2^n)\)
\(C=2^4+2^5+2^6+...+2^{n+1}\)
\(2C-C=\left(2^4+2^5+2^6+...+2^{n+1}\right)-\left(2^3+2^4+2^5+...+2^n\right)\)
\(C=2^{n+1}-2^3\)
Khi đó : \(B=-(2^{n+1}-2^3)+n.2^{n+1}-2^3\)
\(=-2^{n+1}+2^3+n.2^{n+1}-2^3\)
=\(=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n-1}\)
Vậy từ (1) ta có:\(\left(n-1\right),2^{n+1}=2^{n+34}\)
\(2^{n+34}-\left(n-1\right).2^{n+1}=0\)
\(2^{n+1}.[2^{33}-\left(n-1\right)]=0\)
Do đó \(2^{33}-n+1=0\)( Vì \(2^{n+1}\ne0\)với mọi \(n\))
\(n=2^{33}+1\)
Vậy \(n=2^{33}+1\)
21 tháng 11 2016
\(B=\frac{5}{1.2.3}+\frac{5}{2.3.4}+...+\frac{5}{n.\left(n+1\right)\left(n+2\right)}\)
\(\Leftrightarrow\frac{2B}{5}=\frac{2}{1.2.3}+\frac{2}{2.3.4}+...+\frac{2}{n\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+...+\frac{1}{n\left(n+1\right)}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(=\frac{1}{2}-\frac{1}{\left(n+1\right)\left(n+2\right)}\)
\(\Rightarrow B=\frac{5}{4}-\frac{5}{2\left(n+1\right)\left(n+2\right)}\)
PT
1
SM
1