JEE MATHEMATICS MCQ: MATHEMATICAL INDUCTION

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Analyzing +4 | -1

Let

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

is an even integer. If

P\left( k \right)

is assumed true

\Rightarrow P\left( {k + 1} \right)

is true. Therefore,

P\left( n \right)

is true

View answer and explanation

Correct option: D

None of these

Option D (None of these) is correct because the statement P(n) is not true for n=1, thus the principle of mathematical induction cannot be applied.

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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}