If ${x_1} = 3$ and ${x_{n + 1}} = \sqrt {2 + {x_n}} ,\;n \ge 1$, then $\mathop {\lim }\limits_{n \to \infty } {x_n}$ is equal to

JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

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{x_1} = 3

and

{x_{n + 1}} = \sqrt {2 + {x_n}} ,\;n \ge 1

, then

\lim\limits_{n \to \infty } {x_n}

is equal to

View answer and explanation

Correct option: B

Option B (2) is correct because the limit of the recursively defined sequence can be found by solving

x = \sqrt{2+x}

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\frac{1^2}{3}+\frac{1^2+2^2}{5}+\frac{1^2+2^2+3^2}{7}+\ldots

n

P\left( n \right):2 + 4 + 6 \ldots + \left( {2n} \right),\;n \in N,

\begin{array}{l} P\left( k \right) = k\left( {k + 1} \right) + 2\mathrm{implies}\\ P\left( k \right) = \left( {k + 1} \right)\left( {k + 2} \right) + 2 \end{array}

k \in N.

P\left( n \right) = n\left( {n + 1} \right) + 2

n

n

a\left( n \right) = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ... + \frac{1}{{\left( {{2^n}} \right) - 1}}.

n(n+1)(n+2)(n+3)

n \in N

12 {~N}

8 {~m}

6 {~m}