We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
0
172
1
avatar+844 

please explain each step, thanks very much

 Jun 5, 2019
 #1
avatar+23181 
+4

sequence proof

 

\(\text{Let $\mathbf{y=1+x}$ or $\mathbf{x = y-1}$ }\)

\(\begin{array}{|rcll|} \hline f(x) &=& & \dfrac{1}{y} + \dfrac{2}{y^2} + \dfrac{3}{y^3} + \dfrac{4}{y^4} +\ldots + \dfrac{n-1}{y^{n-1}}+ \dfrac{n}{y^n} \quad | \quad \cdot y \\ yf(x) &=&1 +& \dfrac{2}{y } + \dfrac{3}{y^2} + \dfrac{4}{y^3} + \dfrac{5}{y^4} +\ldots + \dfrac{n}{y^{n-1}} \\ \hline yf(x)-f(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \\ (y-1)f(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \quad | \quad y-1 = x \\ xf(x) &=& 1 + & \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} - \dfrac{n}{y^n} \\ \hline \end{array} \)

 

\(\begin{array}{|rclcl|} \hline xf(x) &=& \underbrace{1 + \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}}}_{=s~(\text{sum of a geometric progression})} - \dfrac{n}{y^n} \\ xf(x) &=& \mathbf{s} - \dfrac{n}{y^n} \\ \hline && \mathbf{s} = 1 + \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} \quad | \quad : y \\ && \dfrac{\mathbf{s}}{y} = \quad \dfrac{1}{y } + \dfrac{1}{y^2} + \dfrac{1}{y^3} + \dfrac{1}{y^4} +\ldots + \dfrac{1}{y^{n-1}} + \dfrac{1}{y^{n}} \\ \hline && \mathbf{s}-\dfrac{\mathbf{s}}{y} = 1- \dfrac{1}{y^{n}} \\ && \mathbf{s}\left(1-\dfrac{1}{y}\right) = 1- \dfrac{1}{y^{n}} \\ && \mathbf{s} = \dfrac{1- \dfrac{1}{y^{n}}}{1-\dfrac{1}{y}} \\ \hline xf(x) &=& \mathbf{\dfrac{1- \dfrac{1}{y^{n}}}{1-\dfrac{1}{y}}} - \dfrac{n}{y^n} \quad | \quad y=1+x \\ \mathbf{xf(x)} &=& \mathbf{\dfrac{1- \dfrac{1}{(1+x)^{n}}}{1-\dfrac{1}{1+x}} - \dfrac{n}{(1+x)^n} } \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{xf(x)} &=& \mathbf{\dfrac{1- \dfrac{1}{(1+x)^{n}}}{1-\dfrac{1}{1+x}} - \dfrac{n}{(1+x)^n} } \\ \\ xf(x) &=& \dfrac{ \dfrac{(1+x)^{n}-1}{(1+x)^{n}}} { \dfrac{1+x-1 }{1+x}} - \dfrac{n}{(1+x)^n} \\ \\ xf(x) &=& \dfrac{ \dfrac{(1+x)^{n}-1}{(1+x)^{n}}} { \dfrac{x}{1+x}} - \dfrac{n}{(1+x)^n} \\ \\ xf(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x) } {x(1+x)^{n}} - \dfrac{n}{(1+x)^n}\cdot \dfrac{x}{x} \\ \\ xf(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x)-nx } {x(1+x)^{n}} \quad | \quad : x \\ \\ f(x) &=& \dfrac{\Big( (1+x)^{n}-1 \Big)(1+x)-nx } {x^2(1+x)^{n}} \\ \\ \mathbf{f(x)} &=& \mathbf{ \dfrac{ (1+x)^{n+1}-(1+x)-nx } {x^2(1+x)^{n}} } \\ \hline \end{array}\)

 

laugh

 Jun 5, 2019

15 Online Users

avatar