Tính S theo n ( \(n\in\) N*)
\(S=2^{n-1}+2.2^{n-2}+3.2^{n-3}+...+\left(n-1\right).2+n\)
K
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XO
23 tháng 1 2020
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
ND
0
15 tháng 10 2023
1:
\(K=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^{n+1}+3^{n+1}}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot2^n-3^n}{2^n\cdot2+3^n\cdot3}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3\cdot\dfrac{2^n}{3^n}-1}{\left(\dfrac{2}{3}\right)^n\cdot2+3}\)
\(=-\dfrac{1}{3}\)
2:
\(\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^{n+1}}{3^{n+2}+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{3^n-4^n\cdot4}{3^n\cdot9+4^n}\)
\(=\lim\limits_{n\rightarrow\infty}\dfrac{\left(\dfrac{3}{4}\right)^n-4}{\left(\dfrac{3}{4}\right)^n\cdot9+1}=-\dfrac{4}{1}=-4\)
QN
25 tháng 7 2017
a) \(A=2^{n-1}+2.2^{n+3}-8.2^{n-4}-16.2^n\)
\(=2^{n-1}+2^{n+3+1}-2^{n-4+3}-2^{n+4}\)
\(=2^{n-1}+2^{n+4}-2^{n-1}-2^{n+4}\)
\(=0\)
b) \(B=\left(3^{n+1}-2.2^n\right)\left(3^{n+1}+2.2^n\right)-3^{2n+2}+\left(8.2^{n-2}\right)^2\)
\(=\left(3^{n+1}-2^{n+1}\right)\left(3^{n+1}-2^{n+1}\right)-3^{2n+2}+2^{2n+2}\)
\(=3^{2n+2}-2^{2n+2}-3^{2n+2}+2^{2n+2}\)
\(=0\)
NT
0
Ta có \(2S=2^n+2\cdot2^{n-1}+3\cdot2^{n-2}+...+\left(n-1\right)\cdot2^2+2n\\ \Rightarrow2S-S=2^n+\left(2\cdot2^{n-1}-2^{n-1}\right)+\left(3\cdot2^{n-2}-2\cdot2^{n-2}\right)+...+2n-n\\ \Rightarrow S=2^n+2^{n-1}+2^{n-2}+...+2^2+2-n\\ \Rightarrow S=2\left(2^n-1\right)-n=2^{n+1}-\left(n+2\right)\)
\(S=2^{n-1}+2.2^{n-2}+3.2^{n-3}+...+\left(n-1\right).2+n\)
\(\text{Đặt:}S_n=1.2^{n-1}+2.2^{n-2}+3.2^{n-3}+...+\left(n-1\right).2^1+n\left(1\right)\text{ Với }n\ge1\)
\(\text{Dễ thấy:}S_1=1\)
\(\text{Từ (1) ta có:}\)
\(2S_n+\left(n+1\right)=1.2^n+2.2^{n-1}+3.2^{n-2}+...+\left(n-1\right).2^2+n.2^1+\left(n+1\right)=S_{n+1}\) \(\Rightarrow S_n=2.S_{n-1}+n\)
\(\Leftrightarrow\left(S_n+n+2\right)=2\left(S_{n-1}+\left(n-1\right)+2\right)=2^2\left(S_{n-2}+\left(n-2\right)+2\right)=...=2^{n-1}\left(S_1+\left(1\right)+2\right)=2^{n-1}.4=2^{n+1}\)\(\text{ Do đó ta có:}S_n=2^{n+1}-\left(n+2\right)\)