Prove that:
$a_{n^{2} + 1} = (n^{2} + 1^{2}) - (n^{2} + 1) + 1 = (n^{2} + n + 1) (n^{2} - n + 1) =$

a_{n - 1} a_{n}

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a_{n + 1} a_{n}

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a_{n + 1} a_{n - 1}

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a_{n + 2} a_{n}

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Solution

The correct option is C $a_{n + 1} a_{n}$
Denoting $b_{n} = a_{n + 1} - a_{n},$ we observe that the given recursive equation becomes $b_{n} = b_{n + 1} + 2,$
Thus,
${(b_{n})}_{n \geq 0}$ ,
is an arithmetic sequence and hence $b_{n} = b_{0} + 2 n = 2 n .$
Writing $a_{k + 1} - a_{k} = 2 (k - 1)$ for $k = 1, 2, . . ., n - 1$
and adding up yield $a_{n} = n^{2} - n + 1,$ for all $n \geq 0.$
Now,an elementary computation shows that $a_{n^{2} + 1} = (n^{2} + 1^{2}) - (n^{2} + 1) + 1 = (n^{2} + n + 1) (n^{2} - n + 1) = a_{n + 1} a_{n},$
as desired.

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Similar questions

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

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

a_{n} =

\frac{1}{a_{1} a_{2}} + \frac{1}{a_{2} a_{3}} +

+ \frac{1}{a_{n - 1} a_{n}} = \frac{n - 1}{a_{1} a_{n}}

If $a_{1}, a_{2}, a_{3}, . . . ., a_{n}$ be an AP of non-zero terms. Prove that

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

$\frac{n - 1}{a_{1} a_{n}}$

n^{t h}

a_{n}

a_{1} = \frac{1}{2}

(n - 1) a_{n - 1} = (n + 1) a_{n}

n \geq 2

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