Let a0-1 and an-1-a0- an-4- for all n- 0- Then an-an-1 - for all n- 1-

We have, a_n+1=a_0··· a_n 1· a_n+4 and a_n=a_0··· a_n 1· a_n 1+4 = gt;a_n 4=a_0··· a_n 1· a_n 1 ∴ a_n+1=(a_n 4).a_n+4 = gt;a_n+1=a^2 ...

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Sum of n Terms

Let a0=1 an...

Question

Let $a_{0} = 1$ and $a_{n + 1} = a_{0} \cdot \cdot \cdot a_{n} + 4,$ for all $n \geq 0.$ Then $a_{n} - \sqrt{a_{n + 1}} =$ for all $n \geq 1.$

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(a_{n}) n \geq 0

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a_{0} = 1, a_{1} = 8

a_{10} = ?

a_{0} = 0

a_{n} = 3 a_{n - 1} + 1

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a_{2010}

\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + \dots + \frac{a_{n - 1}}{2} + a_{n} = 0

a_{0} x^{n} + a_{1} x^{n - 1} + \dots + a_{n - 1} x + a_{n} = 0

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\frac{a_{0}}{n + 1} + \frac{a_{1}}{n} + \frac{a_{2}}{n - 1} + . . . . . . + \frac{a_{n - 1}}{2} + a_{n} = 0

a_{0} x^{n} + a_{1} x^{n - 1} + . . . . . . . + . . . . . . + a_{n - 1} x + a_{n} = 0

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