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Đặt A=\(2.2^2+3.2^2+...+n.2^n\)
\(\Rightarrow2A=2.2^3+3.2^4+...+n.2^{n+1}\)
\(\Rightarrow2A-A=\left(2.2^3+3.2^4+...+n.2^{n+1}\right)-\left(2.2^2+3.2^3+...+n.2^n\right)\)
\(\Rightarrow A=n.2^{n+1}-2.2^2-\left(2^3+2^4+...+2^n\right)\)
Đặt \(B=2^3+2^4+...+2^n\Rightarrow2B=2^4+2^5+...+2^{n+1}\)
\(\Rightarrow2B-B=\left(2^4+2^5+...+2^{n+1}\right)-\left(2^3+2^4+...+2^n\right)\)
\(\Rightarrow B=2^{n+1}-2^3\)
\(\Rightarrow A=n.2^{n+1}-2.2^2-\left(2^{n+1}-2^3\right)\)
=\(n.2^{n+1}-8-2^{n+1}+8=\left(n-1\right).2^{n+1}\)
Mà A=\(2^{n+11}\Rightarrow\left(n-1\right).2^{n+1}=2^{n+11}\)
\(\Rightarrow2^{n+1}.\left(n-1\right)=2^{n+1}.2^{10}\Rightarrow n-1=2^{10}=1024\Rightarrow n=2015\)
Vậy...
p hk tốt nha 
\(A=1+2+2^2+2^3+...+2^{10}\)
\(2A=2+2^2+2^3+2^4+...+2^{11}\)
\(2A-A=\left(2+2^2+2^3+2^4+...+2^{11}\right)-\left(1+2+2^2+2^3+...+2^{10}\right)\)
\(A=2^{11}-1< 2^{11}\)
\(B=2.2^2+3.2^3+4.2^4+...+10.2^{10}\)\(2B=2.2^3+3.2^4+4.2^5+...+10.2^{11}\)\(2B-B=\left(2.2^3-3.2^3\right)+\left(3.2^4-4.2^4\right)+...+\left(9.2^{10}-10.2^{10}\right)+10.2^{11}-2.2^2\)\(B=2^3\left(2-3\right)+2^4\left(3-4\right)+...+2^{10}\left(9-10\right)+10.2^{11}-2.2^2\)\(B=-2^3-2^4-....-2^{10}+10.2^{11}-2^3\)
\(B=-\left(2^3+2^4+...+2^{10}\right)+10.2^{11}-2^3\)
\(B=-\left(2^{11}-2^3\right)+10.2^{11}-2^3\)
\(B=-2^{11}+2^3+10.2^{11}-2^3\)
\(B=9.2^{11}\)
Ta cần so sánh: \(9.2^{11}\) và \(2^{14}\)
Hay \(9\) và \(2^3\)
\(9>8=2^3\Leftrightarrow B>2^{14}\)
Có thể không phải là chủ đề Tập hợp Q các số hữu tỉ đâu, nhưng các bạn cứ tự do làm đi.
vì (n2+2) chia hết cho (n+2)
=>n2+2=(n+2)*k
n2+2=k.n+2k
n2-k.n=2k-2
\(a,A=\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-..-\frac{1}{3.2}-\frac{1}{2.1}\)
\(A=\frac{1}{100}-\left(\frac{1}{100.99}-\frac{1}{99.98}-\frac{1}{98.97}-...-\frac{1}{3.2}-\frac{1}{2.1}\right)\)
\(A=\frac{1}{100}-\left(\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{97.98}+\frac{1}{98.99}+\frac{1}{99.100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{97}-\frac{1}{98}+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)
\(A=\frac{1}{100}-1+\frac{1}{100}\)
\(A=\frac{2}{100}-1\)
\(A=\frac{1}{50}-1\)
\(A=\frac{-49}{50}\)
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 \(n=2^{33}+1\)
Đặt A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m Ta có: A = 2.2 2 + 3.2 3 + 4.2 4 + . . . + m .2 m ⇒ 2 A = 2 ( 2.2 2 + 3.2 3 + 4.2 4 + . . . + n .2 n ) ⇒ 2 A = 2.2 3 + 3.2 4 + 4.2 5 + . . . + m .2 m + 1 ⇒ 2 A − A = 2.2 2 + ( 3.2 3 − 2.2 3 ) + . . . + ( m − m + 1 ) .2 m − m .2 m + 1 ⇒ A = 2.2 2 + 2 3 + 2 4 + . . . + 2 n − n .2 n + 1 ⇒ A = 2 2 + ( 2 2 + 2 3 + . . . + 2 m + 1 ) − ( m + 1 ) .2 m + 1 ⇒ A = − 2 2 − ( 2 2 + 2 3 + . . . + 2 m + 1 ) + ( m + 1 ) .2 m + 1 Đặt B = 2 2 + 2 3 + . . . + 2 m + 1 ⇒ 2 B = 2 3 + 2 4 + . . . + 2 m + 2 ⇒ 2 B − B = 2 m + 2 − 2 2 ⇒ B = 2 m + 2 − 2 2 ⇒ A = 2 2 − 2 m + 2 + 2 2 + ( m + 1 ) .2 m + 1 ⇒ A = ( m + 1 ) .2 m + 1 − 2 m + 2 ⇒ A = 2 m + 1 ( m + 1 − 2 ) ⇒ A = ( m − 1 ) .2 m + 1 = 2 ( m − 1 ) .2 n Mà A = 2 ( m − 1 ) .2 m = 2 m + 10 ⇒ 2 ( m + 1 ) = 2 10 ⇒ m − 1 = 2 9 ⇒ m − 1 = 512 ⇒ m = 513 Vậy m = 513