Consider the sequence an given by a1-1-2-an-1-an2-an Let Sn-1-a1-1-1-a2-1-1-an-1 then find the value of -S2012- where - denotes greatest integer function-

a_n+1=a_n^2+a_n=a_n(a_n+1) ⇒1/a_n+1=1/a_n(a_n)+1=1/a_n 1/a_n+1 ⇒1/a_n+1=1/a_n 1/a_n+1⇒1/a_n+1=1/a_n 1/a_n+1 Now S_2012=1/a_1+1+1/a_2+1+1/a_3 ...

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Solving Linear Equations Using Transpose of Terms

Consider the ...

Question

Consider the sequence $a_{n}$ given by $a_{1} = \frac{1}{2}, a_{n} + 1 = a_{n}^{2} + a_{n}$ Let $S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . . . . . . . . . + \frac{1}{a_{n} + 1}$ then find the value of $[S_{2012}]$ , where [.] denotes greatest integer function.

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a_{n}

a_{1} = \frac{1}{2}, a_{n + 1} = a_{n}^{2} + a_{n}

S_{n} = \frac{1}{a_{1} + 1} + \frac{1}{a_{2} + 1} + . . . + \frac{1}{a_{n} + 1}

[S_{2012}]

a_{n}

a_{1} = 1

a_{2} = 1 + a_{1}

a_{3} = 1 + a_{1} a_{2}

a_{n + 1} = 1 + a_{1} a_{2} a_{3} . . . . . . a_{n}

\frac{1}{a_{1}} + \frac{1}{a_{2}} + \frac{1}{a_{3}} + \frac{1}{a_{4}} . . . . . \infty

S_{n} = a_{1} + a_{2} + a_{3} + \dots + a_{n},

a_{1} = 1

a_{n} = 2 a_{n - 1} + 1,

n \geq 2

S_{50}

a_{1} = 0

a_{n + 1} = a_{n} + 4 n + 3

n \geq 1 (n ϵ N)

lim n \to \infty \frac{\sqrt{a_{n}} + \sqrt{a_{9 n}} + \sqrt{a_{9^{2} n}} + . . . . . . + \sqrt{a_{9^{10} n}}}{\sqrt{a_{n}} + \sqrt{a_{3 n}} + \sqrt{a_{3^{2} n}} + . . . . . . + \sqrt{a_{3^{10} n}}} = L,

[\frac{4 L}{3^{11}}]

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