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A = 2.22 + 3.23 + 4.24 + ... + n.2n
2.A = 2.23 + 3.24 + 4.25 + ...+ n.2n+1
=> A - 2.A = 2.22 + (3.23 - 2.23) + (4.24 - 3.24) + ...+ (n - n + 1).2n - n.2n+1
=> A = 2.22 + 23 + 24 + ..+ 2n - n.2n+ 1 = 22 + (22 + 23 + ....+ 2n+ 1) - (n+1).2n+1
=> A = - 22 - (22 + 23 + ....+ 2n+ 1) + (n+1).2n+1
Tính B = 22 + 23 + ....+ 2n+ 1 => 2.B = 23 + ....+ 2n+ 1 + 2n+2 => 2B - B = 2n+2 - 22 => B = 2n+2 - 22
Vậy A = 22 - 2n+2 + 22 + (n+1).2n+1 = (n+1).2n+1 - 2n+ 2 = 2n+1.(n + 1 - 2) = (n-1).2n+1 = 2(n-1).2n
Theo bài cho A = 2(n-1).2n = 2n+10 => 2(n - 1) = 210 => n - 1 = 29 = 512 => n = 513
Vậy.............
Đặt S = 2.22 + 3.23 + 4.24 + ... + (n - 1).2n - 1 + n.2n
<=> S = 2S - S = (2.23 + 3.24 + 4.25 + .... + (n - 1).2n + n. 2n + 1) - (2.22 + 3.23 + 4.24 + ... + (n - 1).2n - 1 + n.2n)
S = (2.23 - 3.23) + (3.24 - 4.24) + (4.25 - 5.25) + .... + [(n - 1).2n - n.2n] + n.2n + 1 - 2.22
= -(23 + 24 + 25 + ... + 2n) + n.2n + 1 - 8
Đặt A = 23 + 24 + 25 + ... + 2n
<=> 2A - A = (24 + 25 + 26 + ... + 2n + 1) - (23 + 24 + 25 + ... + 2n)
<=> A = 2n + 1 - 23
Khi đó S = - 2n - 1 + 23 + n.2n - 1 - 8
= 2n - 1.(n - 1) = 2n + 34
=> n - 1 = 2n + 34 : 2n - 1
=> n - 1 = 2n + 34 - n + 1
=> n - 1 = 235
=> n = 235 + 1
\(\frac{1}{4}\times\frac{2}{6}\times\frac{3}{8}\times...\times\frac{30}{62}\times\frac{31}{64}=2^x\)
\(\Rightarrow\) \(\frac{1\times2\times3\times...\times30\times31}{4\times6\times8\times...\times62\times64}=2^x\)
\(\Rightarrow\frac{1}{2^{31}}=2^x\)\(\Rightarrow1=2^x\times2^{31}\)
\(\Rightarrow2^{x+31}=2^0\)
\(\Rightarrow x+31=0\Rightarrow x=\left(-31\right)\)
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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