Question

Given the sequence series \{a_n\} satisfies a_1=1 , a_2=\dfrac{2}{3} and \dfrac{1}{a_{n+1}} +\dfrac{1}{a_{n-1}}=\dfrac{2}{a_n} (n=2,3,\dots) , find the general term of a_n

Collected in the board: Sequence

Steven Zheng posted 11 months ago


Answer

\because \dfrac{1}{a_{n+1}} +\dfrac{1}{a_{n-1}}=\dfrac{2}{a_n}

\therefore \dfrac{1}{a_{n+1}} -\dfrac{1}{a_n} = \dfrac{1}{a_n}-\dfrac{1}{a_{n-1}}

\dfrac{1}{a_3} -\dfrac{1}{a_2} =\dfrac{1}{a_2} -\dfrac{1}{a_1}

\dfrac{1}{a_4} -\dfrac{1}{a_3} =\dfrac{1}{a_3} -\dfrac{1}{a_2}

\dots

\dfrac{1}{a_{n}} -\dfrac{1}{a_{n-1}} = \dfrac{1}{a_{n-1}}-\dfrac{1}{a_{n-2}}

Add the above equations, we get

\dfrac{1}{a_{n}} -\dfrac{1}{a_2} = \dfrac{1}{a_{n-1}}-\dfrac{1}{a_1}

\therefore \dfrac{1}{a_{n}}-\dfrac{1}{a_{n-1}} = \dfrac{1}{2}

\quad \dfrac{1}{a_2}-\dfrac{1}{a_1} =\dfrac{1}{2}

\quad \dfrac{1}{a_3}-\dfrac{1}{a_2} =\dfrac{1}{2}

\quad \quad \quad\dots

Add above equations, we get,

\quad \dfrac{1}{a_{n}}- \dfrac{1}{a_1} = \dfrac{1}{2} (n-1)

\quad \dfrac{1}{a_{n}} = \dfrac{1}{2} (n+1)

\therefore a_n=\dfrac{2}{n+1}

Steven Zheng posted 11 months ago

Scroll to Top