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1a_1+1a_n=a_n+a_1a_1· a_n 1a_2+1a_n 1=a_n 1+a_2a_2· a_n 1=a_n d+a_1+da_2· a_n 1=a_n+a_1a_2· a_n 1 1a_n+1a_1=a_1+a_na_n· a_1=a_1+a_na_n· a_1 Addin ...

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Show that $\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} +$ ______ $+ \frac{1}{a_{n} a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . + \frac{1}{a_{n}})$ .

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a_{1}, a_{2}, a_{3}, . . . ., a_{n - 1}, a_{n}

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . \frac{1}{a_{n} a_{1}} = \frac{2}{(a_{1} + a_{n})} [\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . . \frac{1}{a_{n}}]

a_{1}, a_{2}, a_{3}, . . . . . . a_{n}

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . + \frac{1}{a_{n} a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + . . . + \frac{1}{a_{n}}) .

a_{1}, a_{2}, a_{3}, . . . . ., a_{n}

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{n - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . + \frac{1}{a_{n} a_{1}} = \frac{2}{a_{1} + a_{n}} (\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + . . . + \frac{1}{a_{n}})

a_{1} + a_{n} = a_{r} + a_{n - r}

1 \leq r \leq n

a_{1}, a_{2}, a_{3} . . . . . a_{n}

\frac{1}{a_{1} a_{n}} + \frac{1}{a_{2} a_{a - 1}} + \frac{1}{a_{3} a_{n - 2}} + . . . . . + \frac{1}{a_{n} a_{1}} =

\frac{a_{1}}{a_{1}} + \frac{a_{2}}{a_{2}} + \frac{a_{3}}{a_{3}} + \dots \frac{a_{n}}{a_{n}} = \frac{1}{1} + \frac{1}{a_{1}} + \frac{a_{1}}{a_{2}} + \frac{a_{2}}{a_{3}} + \dots \frac{a_{n - 2}}{a_{n - 1}}

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