(B) To determine when ${2^n} 1$,the condition is false.For

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(B) To determine when ${2^n} For $n = 1$: ${2^1} = 2$,$1! = 1$. Since $2 > 1$,the condition is false.For $n = 2$: ${2^2} = 4$,$2! = 2$. Since $4 > 2$,the condition is false.For $n = 3$: ${2^3} = 8$,$3! = 6$. Since $8 > 6$,the condition is false.For $n = 4$: ${2^4} = 16$,$4! = 24$. Since $16 For $n = 5$: ${2^5} = 32$,$5! = 120$. Since $32 Thus,the inequality ${2^n} < n!$ holds for all integers $n \geq 4$.

Class 11 Mathematics Mathematical induction Mathematical induction

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For every positive integer $n$,${2^n} < n!$ when

A
$n < 4$
B
$n \geq 4$
C
$n < 3$
D
None of these

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Mathematical induction

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Mathematical induction

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Prove that the following is true for all $n \in N$ using the principle of mathematical induction:
$n(n+1)(n+5)$ is a multiple of $3.$

Difficult

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For every positive integer $n$,${2^n} < n!$ when

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Prove that the following is true for all $n \in N$ using the principle of mathematical induction:
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