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25 tháng 12 2023

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

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 haha

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 \(B=2^2+2^3+...+2^{n+1}\)

\(\Rightarrow2B=2^3+2^4+...+2^{n+2}\)

\(\Rightarrow2B-B=2^{n+2}-2^2\Rightarrow B=2^{n+2}-2^2\)

\(\Rightarrow A=2^2-2^{n+2}+2^2+\left(n+1\right).2^{n+1}\)

\(\Rightarrow A=\left(n+1\right).2^{n+1}-2^{n+2}\)

\(\Rightarrow A=2^{n+1}\left(n+1-2\right)\)

\(\Rightarrow A=\left(n-1\right).2^{n+1}=2\left(n-1\right).2^n\)

\(A=2\left(n-1\right).2^n=2^{n+10}\)

\(\Rightarrow2\left(n+1\right)=2^{10}\Rightarrow n-1=2^9\)

\(\Rightarrow n-1=512\Rightarrow n=513\)

Vậy \(n=513\)

31 tháng 3 2017

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8 tháng 4 2017

Bạn tham khảo ở đay nha:

Câu hỏi của Thái Hoàng Thục Anh - Toán lớp 7 - Học toán với OnlineMath

14 tháng 3 2019

Đặt \(A=2.2^2+3.2^3+4.2^4+...+n.2^n\)

\(\Leftrightarrow2A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}\)

\(\Leftrightarrow2A-A=2.2^3+3.2^4+4.2^5+...+n.2^{n+1}-\left(2.2^2+3.2^3+4.2^4+...+n.2^n\right)\)

\(\Leftrightarrow A=-2.2^2-2^3-2^4-....-2^n+n.2^{n+1}\)

\(\Leftrightarrow A=-2^{n+1}+n.2^{n+1}=\left(n-1\right).2^{n+1}\)

\(A=2^{n+11}\) \(\Leftrightarrow\left(n-1\right).2^{n+1}=2^{n+11}\)

\(\Leftrightarrow\left(n-1\right).2^n.2=2^n.2^{11}\)

\(\Leftrightarrow\left(n-1\right)=2^{10}\)

\(\Leftrightarrow n=2^{10}+1\)

10 tháng 7 2017

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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