JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

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a_n=\sqrt{7+\sqrt{7+\sqrt{7+\ldots \ldots}}}

having n radical signs then by methods of mathematical induction which is true

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Correct option: B

a_n < 7 \,\forall\, n \geq 1

Option B (

a_n < 7 \forall n \geq 1

) is correct because by mathematical induction, it's shown that each term is less than 7.

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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}