Giải bài 3.48 trang 134 - SBT Đại số và Giải tích lớp 11
Tính tổng
a) \(S_n=1+2a+3a^2+...+na^{n-1}\)
b) \(S_n=1.x+2.x^2+3.x^3+...+n.x^n\)
a)
Với \(a=1\) ta có:
\({{S}_{n}}=1+2+3+...+n=\dfrac{n\left( n+1 \right)}{2} \)
Với \(a\ne 1\) ta có:
\(\begin{aligned} & a{{S}_{n}}=a+2{{a}^{2}}+3{{a}^{3}}+...+n{{a}^{n}} \\ & \Rightarrow {{S}_{n}}-a{{S}_{n}}=\left( 1-a \right){{S}_{n}} \\ & =\left( 1+2a+3{{a}^{2}}+...+n{{a}^{n-1}} \right)-\left( a+2{{a}^{2}}+3{{a}^{3}}+...+n{{a}^{n}} \right) \\ & =1+a+{{a}^{2}}+...+{{a}^{n-1}}-n{{a}^{n}} \\ & =\dfrac{{{a}^{n}}-1}{a-1}-n{{a}^{n}} \\ & =\dfrac{{{a}^{n}}-1-n{{a}^{n+1}}+n{{a}^{n}}}{a-1} \\ & =\dfrac{\left( 1+n \right){{a}^{n}}-n{{a}^{n+1}}-1}{a-1} \\ & \Rightarrow {{S}_{n}}=\dfrac{n{{a}^{n+1}}-\left( n+1 \right){{a}^{n}}+1}{{{\left( a-1 \right)}^{2}}} \\ \end{aligned} \)
b) Với \(x=1\) ta có: \({{S}_{n}}=1+2+3+...+n=\dfrac{n\left( n+1 \right)}{2} \)
Với \(x\ne 1\) ta có:
\(\begin{aligned} & x{{S}_{n}}=1.{{x}^{2}}+2.{{x}^{3}}+3{{x}^{4}}+...+n{{x}^{n+1}} \\ & \Rightarrow {{S}_{n}}-x{{S}_{n}}=\left( 1-x \right){{S}_{n}} \\ & =\left( 1.x+2{{x}^{2}}+3{{x}^{3}}+...+n{{x}^{n}} \right)-\left( 1.{{x}^{2}}+2{{x}^{3}}+3{{x}^{4}}+...+n{{x}^{n+1}} \right) \\ & =x+{{x}^{2}}+{{x}^{3}}+...+{{x}^{n}}-n{{x}^{n+1}} \\ & =\dfrac{x\left( {{x}^{n}}-1 \right)}{x-1}-n{{x}^{n+1}} \\ & =\dfrac{{{x}^{n+1}}-x-n{{x}^{n+2}}+n{{x}^{n+1}}}{x-1} \\ & =\dfrac{\left( n+1 \right){{x}^{n+1}}-n{{x}^{n+2}}-x}{x-1} \\ & \Rightarrow {{S}_{n}}=\dfrac{n{{x}^{n+2}}-\left( n+1 \right){{x}^{n+1}}+x}{{{\left( x-1 \right)}^{2}}} \\ \end{aligned} \)