Tìm m thoả mãn điều kiện sau (m ϵ N)
2.2\(^2\)+3.2\(^3\)+\(4.2^4\)+.....+m.2\(^m\) = \(2^{m+11}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
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
2
31 tháng 3 2017
Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
Ta có:
\(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(\Rightarrow2A=2\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(\Rightarrow2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
\(\Rightarrow2A-A=2.2^2+\left(3.2^3-2.2^3\right)+...+\left(n-n+1\right).2^n-n.2^{n+1}\)
\(\Rightarrow A=2.2^2+2^3+2^4+...+2^n-n.2^{n+1}\)
\(\Rightarrow A=2^2+\left(2^2+2^3+...+2^{n+1}\right)-\left(n+1\right).2^{n+1}\)
\(\Rightarrow A=-2^2-\left(2^2+2^3+...+2^{n+1}\right)+\left(n+1\right).2^{n+1}\)
Đặt \(B=2^2+2^3+...+2^{n+1}\)
\(\Rightarrow2B=2^3+2^4+...+2^{n+2}\)
\(\Rightarrow2B-B=2^{n+2}-2^2\Rightarrow B=2^{n+2}-2^2\)
\(\Rightarrow A=2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)
\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)
\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)
\(\Rightarrow A=\left(n-1\right).2^{n+1}=2\left(n-1\right).2^n\)
Mà \(A=2\left(n-1\right).2^n=2^{n+10}\)
\(\Rightarrow2\left(n+1\right)=2^{10}\Rightarrow n-1=2^9\)
\(\Rightarrow n-1=512\Rightarrow n=513\)
Vậy \(n=513\)
31 tháng 3 2017
Bạn vào câu hỏi tương tự hoặc link https://hoc24.vn/hoi-dap/question/213510.html nha 
NM
1
NT
2
NT
Nguyễn Thị Thương Hoài
Giáo viên
VIP
17 tháng 1
2.22 + 3.23 + 4.24 + ... + n.2n = 2n+11
Đặt vế trái là A ta có:
A = 2.22 + 3.23 + 4.24 + ... (n -1).2n-1+ n.2n
2A = 2.23 + 3.24 + 4.25 +....+ (n- 1).2n + n.2n+1
2A-A = [2.23+3.24 + 4.25 +...+(n-1).2n+n.2n+1] - [2.22 + 3.23+...+n.2n]
A = -2.22+ (2.33-3.23) + (3.24 - 4.24) +...+ [(n-1).2n - n.2n ] + n.2n+1
A = -2.22 - 23 - 24 -...- 2n + n.2n+1
Đặt B = -2.22 - 23 - 24 - ... - 2n
2B = -2.23 - 24 - 25 -...-2n+1
2B - B = (-2.23 - 24 - 25 -..-2n+1) - (-2.22-23-24-..-2n)
B = -24 -24 - 25 -..2n-2n+1 + 23 + 23 + 24+ 25+ 2n
B = (-24 + 23) + (- 2n+1 + 23) +(-24+24)+(-25+25)+(-2n+2n)
B = -16 + 8 - 2n+1 + 8
B = (-16 + 8 + 8 ) - 2n+1
B = - 2n+1
A = n.2n+1 - 2n+1
Theo bài ra ta có:
n.2n+1 - 2n+1 = 2n+11
n.2n+1 - 2n+1 - 2n+11 = 0
2n+1.(n - 1 - 210) = 0
Vì n là số tự nhiên nên 2n+1 > 0
Vậy 2n+1.(n - 1- 210) = 0 ⇔ n - 1 - 210 = 0 ⇒ n = 1 + 210 ⇒ n = 1025
Vậy n = 1025
AW
2
14 tháng 3 2019
Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)
\(\Leftrightarrow2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)
\(\Leftrightarrow2A-A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)
\(\Leftrightarrow A=-2.2^2-2^3-2^4-....-2^n+n.2^{n+1}\)
\(\Leftrightarrow A=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n+1}\)
mà \(A=2^{n+11}\) \(\Leftrightarrow\left(n-1\right).2^{n+1}=2^{n+11}\)
\(\Leftrightarrow\left(n-1\right).2^n.2=2^n.2^{11}\)
\(\Leftrightarrow\left(n-1\right)=2^{10}\)
\(\Leftrightarrow n=2^{10}+1\)
D
0
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
VT
0
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