If a1 - 1- an-1 - 1n-1an- n - 1- then an -

If $a_{1}, a_{2}, a_{3} . . . . . a_{n}$ are in A.P. Where $a_{i} > 0$ for all i, then the value of
$\frac{1}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{1}{\sqrt{a_{2}} + \sqrt{a_{3}}} + . . . . . . . . . + \frac{1}{\sqrt{a_{n - 1}} + \sqrt{a_{n}}} =$

Let a_{1} = 2, a_{n + 1} = a_{n}^{2} - a_{n} + 1 for n \geq 1, then \infty \sum n = 1 \frac{1}{a_{n}} is

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If a1=1,an+1=1n+1an,∀n≥1, then an =

The correct option is A 1n!a1=1,a2=11+1a1=12(1)=12!,a3=13.12!=13!Similarly an=1n.1(n−1)!=1n!

If $a_{1} = 1, a_{n + 1} = \frac{1}{n + 1} a_{n}, \forall n \geq 1$ , then $a_{n} =$

The correct option is A $\frac{1}{n!}$
$a_{1} = 1, a_{2} = \frac{1}{1 + 1} a_{1} = \frac{1}{2} (1) = \frac{1}{2!}, a_{3} = \frac{1}{3} . \frac{1}{2!} = \frac{1}{3!}$
Similarly $a_{n} = \frac{1}{n} . \frac{1}{(n - 1)!} = \frac{1}{n!}$