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
231
1
avatar+283 

deleted

 Jul 15, 2018
edited by yasbib555  Jul 17, 2018
 #1
avatar+22550 
+2

Given {a_n}, a_n=4/(2n-1)^2 and given {bn}, bn=(1-a_1)(1-a_2)...(1-a_n),

please use mathematical induction to prove that b_n=(2n+1)/(1-2n ). 

\(\begin{array}{|rcll|} \hline a_n &=& \dfrac{4}{(2n-1)^2} \\\\ b_n &=& (1-a_1)(1-a_2)\ldots (1-a_n) \\\\ b_n &=& \dfrac{2n+1}{1-2n} \\ \hline \end{array}\)

 

\(\mathbf{n = 1}\)

\(\begin{array}{|rcll|} \hline a_1 &=& \dfrac{4}{(2\cdot 1-1)^2} \\\\ &=& \dfrac{4}{1} \\\\ &=& 4 \\\\ b_1 &=& \dfrac{2\cdot 1+1}{1-2\cdot 1} \\\\ &=& \dfrac{3}{-1} \\\\ &=& -3 \\\\ b_1 &=& (1-a_1) \\\\ &=& 1-4 \\ &=&\ -3\ \checkmark \\ \hline \end{array} \)

 

\(\mathbf{n = k}\)

\(\begin{array}{|rcll|} \hline a_k &=& \dfrac{4}{(2k-1)^2} \\\\ b_k &=& (1-a_1)(1-a_2)\ldots (1-a_k) \\\\ b_k &=& \dfrac{2k+1}{1-2k} \qquad \mathbf{b_{k+1} = \dfrac{2(k+1)+1}{1-2(k+1)}} \\ \hline \end{array}\)

 

\(\mathbf{k+1} \)

\(\begin{array}{|rcll|} \hline a_{k+1} &=& \dfrac{4}{(2(k+1)-1)^2} \\ &=& \dfrac{4}{(2k+2-1)^2} \\ &=& \dfrac{4}{(2k+1)^2} \\\\ b_{k+1} &=& (1-a_1)(1-a_2)\ldots (1-a_k)(1-a_{k+1}) \\ &=& b_k(1-a_{k+1}) \\ &=& b_k \left(1- \dfrac{4}{(2k+1)^2} \right) \quad & | \quad b_k = \dfrac{2k+1}{1-2k} \\\\ &=& \left(\dfrac{2k+1}{1-2k}\right) \left(1- \dfrac{4}{(2k+1)^2} \right) \\\\ &=& \left(\dfrac{2k+1}{1-2k}\right) \left( \dfrac{(2k+1)^2-4}{(2k+1)^2} \right) \\\\ &=& \dfrac{\left(2k+1\right)\Big( (2k+1)^2-4 \Big)}{\left(1-2k\right)(2k+1)^2} \\\\ &=& \dfrac{(2k+1)^2-4}{\left(1-2k\right)(2k+1)} \\\\ &=& \dfrac{4k^2+4k+1-4}{\left(1-2k\right)(2k+1)} \\\\ &=& \dfrac{4k^2+4k-3}{(1-2k)(2k+1)} \\\\ &=& -\dfrac{(2k+3)(1-2k)}{(1-2k)(2k+1)} \\\\ &=& \dfrac{(2k+3)}{-(2k+1)} \\\\ &=& \dfrac{2k+3}{-2k-1} \\\\ &\mathbf{=}& \mathbf{\dfrac{2(k+1)+1}{1-2(k+1)}}\ \checkmark \\ \hline \end{array}\)

 

laugh

 Jul 16, 2018
edited by heureka  Jul 17, 2018

11 Online Users

avatar