NM
1
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Các câu hỏi dưới đây có thể giống với câu hỏi trên
VT
0
OS
0
NT
2
TN
1
26 tháng 3 2017
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 
NM
0
NM
2
10 tháng 7 2017
https://h.vn/hoi-dap/question/221389.html kham khảo ak!!! bài dài quá lười đánh máy lắm, thông cảm!!!^~
LL
27 tháng 3 2019
Đặt A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n
Ta có:
A=2.22+3.23+4.24+...+n.2nA=2.22+3.23+4.24+...+n.2n
⇒2A=2(2.22+3.23+4.24+...+n.2n)⇒2A=2(2.22+3.23+4.24+...+n.2n)
⇒2A=2.23+3.24+4.25+...+n.2n+1⇒2A=2.23+3.24+4.25+...+n.2n+1
⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1⇒2A−A=2.22+(3.23−2.23)+...+(n−n+1).2n−n.2n+1
⇒A=2.22+23+24+...+2n−n.2n+1⇒A=2.22+23+24+...+2n−n.2n+1
⇒A=22+(22+23+...+2n+1)−(n+1).2n+1⇒A=22+(22+23+...+2n+1)−(n+1).2n+1
⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1⇒A=−22−(22+23+...+2n+1)+(n+1).2n+1
Đặt B=22+23+...+2n+1B=22+23+...+2n+1
⇒2B=23+24+...+2n+2⇒2B=23+24+...+2n+2
⇒2B−B=2n+2−22⇒B=2n+2−22⇒2B−B=2n+2−22⇒B=2n+2−22
⇒A=22−2n+2+22+(n+1).2n+1⇒A=22−2n+2+22+(n+1).2n+1
⇒A=(n+1).2n+1−2n+2⇒A=(n+1).2n+1−2n+2
⇒A=2n+1(n+1−2)⇒A=2n+1(n+1−2)
⇒A=(n−1).2n+1=2(n−1).2n⇒A=(n−1).2n+1=2(n−1).2n
Mà A=2(n−1).2n=2n+10A=2(n−1).2n=2n+10
⇒2(n+1)=210⇒n−1=29⇒2(n+1)=210⇒n−1=29
⇒n−1=512⇒n=513⇒n−1=512⇒n=513
Vậy n=513
NH
2
LC
25 tháng 10 2015
Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
=>\(2.A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
=>\(A-2A=2.2^2+3.2^3+4.2^4+...+n.2^n-2.2^3-3.2^4-4.2^5-...-n.2^{n+1}\)
=>\(-A=2.2^2+\left(3.2^3-2.2^3\right)+\left(4.2^4-3.2^4\right)+...+\left(n.2^n-\left(n-1\right).2^n\right)-n.2^{n+1}\)
=>\(-A=2^3+2^3+2^4+...+2^n-n.2^{n+1}\)
=>\(-A=2^3+\left(2^3+2^4+...+2^n\right)-n.2^{n+1}\)
=>\(A=n.2^{n+1}-2^3-\left(2^3+2^4+...+2^n\right)\)
Đặt \(B=2^3+2^4+...+2^n\)
=>\(2.B=2^4+2^5+...+2^{n+1}\)
=>\(2.B-B=2^4+2^5+...+2^{n+1}-2^3-2^4-...-2^n\)
=>\(B=2^{n+1}-2^3\)
Lại có:\(A=n.2^{n+1}-2^3-\left(2^3+2^4+...+2^n\right)\)
=>\(A=n.2^{n+1}-2^3-B\)
=>\(A=n.2^{n+1}-2^3-\left(2^{n+1}-2^3\right)\)
=>\(A=n.2^{n+1}-2^3-2^{n+1}+2^3\)
=>\(A=n.2^{n+1}-2^{n+1}\)
=>\(A=\left(n-1\right).2^{n+1}\)
Mà \(A=2.2^2+3.2^3+4.2^4+...+n.2^n=2^{n+10}\)
=>\(\left(n-1\right).2^{n+1}=2^{n+10}\)
=>\(n-1=2^{n+10}:2^{n+1}\)
=>\(n-1=2^{n+10-n-1}\)
=>\(n-1=2^9\)
=>\(n-1=512\)
=>\(n=513\)
Vậy n=513
25 tháng 10 2015
dài thế hình như cô giáo lớp mình giải còn ngắn hơn thế này
NH
0
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 \(n=2^{33}+1\)
\(Đặt\) \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(2A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}\)
\(2A-A=2.2^3+3.2^4+4.2^5+....+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(=-2.2^2-2^3-2^4-...-2^n+n.2^{n+1}\)
\(=-2^2-\left(2^2+2^3+...+2^n\right)+n.2^{n+1}\)
\(=-2^2-\left(2^{n+1}-2^2\right)+n.2^{n+1}\)
\(=\left(n-1\right).2^{n+1}\)
=> \(\left(n-1\right).2^{n+1}=2^{n+16}=2^{n+1}.2^{15}\)
\(\Leftrightarrow n-1=2^{15}\)
\(\Leftrightarrow n=2^{15}+1\)