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Q)

If $f_{1}=f_{2}=1$ and thereafter $f_{n+2}=f_{n+1}+f_{n}$ for all $n \in N$. Find $\displaystyle\sum_{n=2}^{\infty}\;\large\frac{1}{f_{n+1}\;.f_{n-1}}$

$(a)\;2\qquad(b)\;1\qquad(c)\;\large\frac{1}{2}\qquad(d)\;None\;of\;these$

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A)
Answer : (b) 1
Explanation : $\;Let \;S=\displaystyle\sum_{n=2}^{\infty}\;\large\frac{1}{f_{n+1}.f_{n-1}}$
$Let \;S=\displaystyle\sum_{n=2}^{\infty}\;\large\frac{f_{n}}{(f_{n+1}.f_{n-1})f_{n}}$
$f_{n+1}=f_{n}+f_{n-1}$
$f_{n}=f_{n+1}-f_{n-1}$
$S=\displaystyle\sum_{n=2}^{\infty}\;\large\frac{f_{n+1}-f_{n-1}}{(f_{n+1}.f_{n-1})f_{n}}$
$S=\displaystyle\sum_{n=2}^{\infty}\;\large\frac{1}{f_{n}f_{n-1}}-\large\frac{1}{f_{n}f_{n+1}}$
$S=\large\frac{1}{1.1}=1\;.$
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