Consider the statement: $P(n): n^2 - n + 41$ is prime. Then which one of the following is true?
- ABoth $P(3)$ and $P(5)$ are true
- B$P(3)$ is false but $P(5)$ is true
- CBoth $P(3)$ and $P(5)$ are false
- D$P(5)$ is false but $P(3)$ is true
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Similar Questions
Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$
Medium
View SolutionUsing mathematical induction,the numbers $a_n$ are defined by:
$a_0 = 1, a_{n+1} = 3n^2 + n + a_n, (n \geq 0)$.
Then,$a_n$ is equal to:
$a_0 = 1, a_{n+1} = 3n^2 + n + a_n, (n \geq 0)$.
Then,$a_n$ is equal to:
DifficultAP EAMCET 2009
View SolutionFor all natural numbers $n$,$3(5^{2n+1}) + 2^{3n+1}$ is divisible by:
Prove the following by using the principle of mathematical induction for all $n \in N:$
$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$
$\frac{1}{1 \cdot 3} + \frac{1}{3 \cdot 5} + \frac{1}{5 \cdot 7} + \ldots + \frac{1}{(2n-1)(2n+1)} = \frac{n}{2n+1}$
Medium
View SolutionUse the Principle of Mathematical Induction to prove that $\cos \theta \cos 2 \theta \cos 2^{2} \theta \ldots \cos 2^{n-1} \theta = \frac{\sin 2^{n} \theta}{2^{n} \sin \theta}$ for all $n \in N$.
Difficult
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