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yasbib555  Jul 15, 2018
edited by yasbib555  Jul 17, 2018
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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

heureka  Jul 16, 2018
edited by heureka  Jul 17, 2018

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