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a) Ta có:
\(q.{S_n} = q.\left( {{u_1} + {u_2} + ... + {u_n}} \right) = {u_1}.q + {u_2}.q + ... + {u_n}.q = \left( {{u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n}\)
b) Ta có:
\({u_1} + q.{S_n} = {u_1} + \left( {{u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n} = \left( {{u_1} + {u_2} + {u_3} + ... + {u_n}} \right) + q.{u_n} = {S_n} + {u_1}.{q^n}\)
a) \(\left| q \right| = \left| {\frac{1}{2}} \right| < 1\)
b) \(\begin{array}{l}{S_n} = {u_1} + {u_2} + ... + {u_n} = {u_1}.\frac{{1 - {q^n}}}{{1 - q}} = 1.\frac{{1 - {{\left( {\frac{1}{2}} \right)}^n}}}{{1 - \frac{1}{2}}} = 2 - 2.{\left( {\frac{1}{2}} \right)^n}\\ \Rightarrow \lim {S_n} = \lim \left[ {2 - 2.{{\left( {\frac{1}{2}} \right)}^n}} \right] = \lim 2 - 2\lim {\left( {\frac{1}{2}} \right)^n} = 2\end{array}\)
a) Ta có:
\(\left. \begin{array}{l}{u_1} + {u_n} = {u_1} + {u_1} + \left( {n - 1} \right)d = 2{u_1} + \left( {n - 1} \right)d\\{u_2} + {u_{n - 1}} = {u_1} + d + \left( {n - 2} \right)d = {u_1} + \left( {n - 1} \right)d\\{u_n} + {u_1} = {u_1} + {u_1} + \left( {n - 1} \right)d = 2{u_1} + \left( {n - 1} \right)d\end{array} \right\} \Rightarrow {u_1} + {u_n} = {u_2} + {u_{n - 1}} = ... = {u_n} + {u_1}\)
b) Dựa vào công thức vừa chứng minh ta có: \(n\left( {{u_1} + {u_n}} \right)\) = \(2{S_n}\)
a) \({u_2} = {u_1}.q\)
\({u_3} = {u_1}.{q^2}\)
…
\({u_{n - 1}} = {u_1}.{q^{n - 2}}\)
\({u_n} = {u_1}.{q^{n - 1}}\)
\({S_n} = {u_1} + {u_1}q + \ldots + {u_1}{q^{n - 2}} + {u_1}{q^{n - 1}}\)
b) \(q{S_n} = q{u_1} + {u_1}{q^2} + \ldots + {u_1}{q^{n - 1}} + {u_1}{q^n}\)
c) \({S_n} - q{S_n} = \left( {{u_1} + {u_1}q + \ldots + {u_1}{q^{n - 2}} + {u_1}{q^{n - 1}}} \right) - (q{u_1} + {u_1}{q^2} + \ldots + {u_1}{q^{n - 1}} + {u_1}{q^n})\).
\(\begin{array}{l} \Leftrightarrow \left( {1 - q} \right){S_n} = {u_1} - {u_1}{q^n} = {u_1}\left( {1 - {q^n}} \right)\\ \Rightarrow {S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{1 - q}}\end{array}\)
a) \({u_2} = {u_1}.q\)
\({u_3} = {u_2}.q = {u_1}.{q^2}\)
\({u_4} = {u_3}.q = {u_1}.{q^3}\)
\({u_5} = {u_4}.q = {u_1}.{q^4}\)
b) Từ a suy ra: \({u_n} = {u_1} \times {q^{n - 1}}\).
a) \({u_2} = {u_1} + d\)
\({u_3} = {u_1} + 2d\)
…
\({u_{n - 1}} = {u_1} + \left( {n - 2} \right)d\)
\({u_n} = {u_1} + \left( {n - 1} \right)d\)
\({S_n} = {u_1} + {u_1} + 2d + \ldots + {u_1} + \left( {n - 2} \right)d + {u_1} + \left( {n - 1} \right)d\)
b) \({S_n} = {u_n} + {u_{n - 1}} + \ldots + {u_2} + {u_1} = {u_1} + \left( {n - 1} \right)d + {u_1} + \left( {n - 2} \right)d + \ldots + {u_1} + d + {u_1}\)
c) \(2{S_n} = \left( {{u_1} + {u_1} + d + \ldots + {u_1} + \left( {n - 1} \right)d} \right) + \left( {{u_1} + \left( {n - 1} \right)d + {u_1} + \left( {n - 2} \right)d + \ldots + {u_1}} \right)\).
\( \Rightarrow 2{S_n} = n.\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)
\( \Rightarrow {S_n} = \frac{n}{2}\left( {2{u_1} + \left( {n - 1} \right)d} \right)\)
Câu 1:
\(S_8=u_1+u_2+u_3+...+u_8\)
\(=\dfrac{u_1\left(1-q^8\right)}{1-q}=\dfrac{2048\cdot\left(1-\left(\dfrac{5}{4}\right)^8\right)}{1-\dfrac{5}{4}}\)
\(=\dfrac{325089}{8}\)
2: \(S_{10}=u_1+u_2+...+u_9+u_{10}\)
=>\(S_{10}=\dfrac{u_1\left(1-q^{10}\right)}{1-q}=\dfrac{-3\cdot\left(1-\left(\dfrac{1}{2}\right)^{10}\right)}{1-\dfrac{1}{2}}\)
\(=-6\cdot\left(1-\dfrac{1}{2^{10}}\right)=-6+\dfrac{6}{2^{10}}=-\dfrac{3069}{512}\)
\(u_3=u_2^2-u_2+2=4\)
\(S_1=1=\left(2-1\right)^2=\left(u_2-1\right)^2\)
\(S_2=2.5-1=9=\left(4-1\right)^2=\left(u_3-1\right)^2\)
Dự đoán \(S_n=\left(u_{n+1}-1\right)^2\)
Ta sẽ chứng minh bằng quy nạp:
- Với \(n=1;2\) đúng (đã kiểm chứng bên trên với \(S_1;S_2\))
- Giả sử đẳng thức đúng với \(n=k\)
Hay \(S_k=\left(u_1^2+1\right)\left(u_2^2+1\right)...\left(u_k^2+1\right)-1=\left(u_{k+1}-1\right)^2\)
Ta cần chứng minh:
\(S_{k+1}=\left(u_1^2+1\right)\left(u_2^2+1\right)...\left(u_k^2+1\right)\left(u_{k+1}^2+1\right)-1=\left(u_{k+2}-1\right)^2\)
Thật vậy:
\(S_{k+1}=\left[\left(u_{k+1}-1\right)^2+1\right]\left(u_{k+1}^2+1\right)-1\)
\(=\left(u_{k+1}^2-2u_{k+1}+2\right)\left(u_{k+1}^2+1\right)-1\)
\(=\left(u_{k+2}-u_{k+1}\right)\left(u_{k+2}+u_{k+1}-1\right)-1\)
\(=u_{k+2}^2-u_{k+2}-u_{k+1}^2+u_{k+1}-1\)
\(=u_{k+2}^2-u_{k+2}+2-u_{k+2}-1\)
\(=\left(u_{k+2}-1\right)^2\) (đpcm)
a) Ta có:
\({S_n}.q = \left( {{u_1} + {u_1}q + {u_1}{q^2} + ... + {u_1}{q^{n - 1}}} \right).q = {u_1}\left( {1 + q + {q^2} + ... + {q^{n - 1}}} \right).q = {u_1}\left( {q + {q^2} + {q^3} + ... + {q^n}} \right)\)
\(\begin{array}{l}{S_n} - {S_n}.q = {u_1} + {u_1}q + {u_1}{q^2} + ... + {u_1}{q^{n - 1}} - {u_1}\left( {q + {q^2} + {q^3} + ... + {q^n}} \right)\\ = {u_1}\left( {1 + q + {q^2} + ... + {q^{n - 1}}} \right) - {u_1}\left( {q + {q^2} + {q^3} + ... + {q^n}} \right)\\ = {u_1}\left( {1 + q + {q^2} + ... + {q^{n - 1}} - \left( {q + {q^2} + {q^3} + ... + {q^n}} \right)} \right)\\ = {u_1}\left( {1 - {q^n}} \right)\end{array}\)
b) Ta có: \({S_n} - {S_n}.q = {u_1}\left( {1 - {q^n}} \right) \Leftrightarrow {S_n}\left( {1 - q} \right) = {u_1}\left( {1 - {q^n}} \right) \Leftrightarrow {S_n} = \frac{{{u_1}\left( {1 - {q^n}} \right)}}{{\left( {1 - q} \right)}}\)