Toán 2.2^2+3.2^3+4.2^4+…+(n-1).2^n-1+n.2^n=2^n+34 tìm n 08/10/2021 By Rose 2.2^2+3.2^3+4.2^4+…+(n-1).2^n-1+n.2^n=2^n+34 tìm n
Đáp án: 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 Giải thích các bước giải: 10 điểm Trả lời
Giải thích các bước giải: Đặt `A=2.2^2+3.2^3+4.2^4+…+(n-1).2^(n-1)+n.2^n` `=> 2A = 2.2^3 + 3.2^4 + 4.2^5 + … + (n-1).2^n + n.2^(n+1)` `=> 2A-A= n . 2^(n+1) – 2^3 – (2^3 + 2^4 + …. + 2^(n-1) + 2^n)` Đặt `B = 2^3 + 2^4 + …. + 2^(n-1) + 2^n` `=> 2B = 2^4 + 2^5 + … + 2^n + 2^(n+1)` `=> 2B – B = (2^4 + 2^5 + … + 2^n + 2^(n+1)) – (2^3 + 2^4 + …. + 2^(n-1) + 2^n)` `=> B = 2^(n+1) – 2^3` `=> A = n.2^(n+1) – 2^3 – 2^3 – (2^(n+1) – 2^3)` `= n.2^(n+1) – 2^3 – 2^(n+1) + 2^3` `= n.2^(n+1) – 2^(n+1) = 2^(n+1) . (n-1)` `=> 2^(n+1) ( n-1) = 2^(n+34)` `=> n-1= 2^(n+34) : 2^(n+1)` `=> n-1 = 2^33` `=> n = 2^33 + 1` Trả lời
Đáp án:
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
Giải thích các bước giải:
10 điểm
Giải thích các bước giải:
Đặt `A=2.2^2+3.2^3+4.2^4+…+(n-1).2^(n-1)+n.2^n`
`=> 2A = 2.2^3 + 3.2^4 + 4.2^5 + … + (n-1).2^n + n.2^(n+1)`
`=> 2A-A= n . 2^(n+1) – 2^3 – (2^3 + 2^4 + …. + 2^(n-1) + 2^n)`
Đặt `B = 2^3 + 2^4 + …. + 2^(n-1) + 2^n`
`=> 2B = 2^4 + 2^5 + … + 2^n + 2^(n+1)`
`=> 2B – B = (2^4 + 2^5 + … + 2^n + 2^(n+1)) – (2^3 + 2^4 + …. + 2^(n-1) + 2^n)`
`=> B = 2^(n+1) – 2^3`
`=> A = n.2^(n+1) – 2^3 – 2^3 – (2^(n+1) – 2^3)`
`= n.2^(n+1) – 2^3 – 2^(n+1) + 2^3`
`= n.2^(n+1) – 2^(n+1) = 2^(n+1) . (n-1)`
`=> 2^(n+1) ( n-1) = 2^(n+34)`
`=> n-1= 2^(n+34) : 2^(n+1)`
`=> n-1 = 2^33`
`=> n = 2^33 + 1`