Let $P(n) = 3^{2n+1} + 2^{n+2}$ where $n \in N$. Then
- A$P(n)$ is not divisible by any prime integer.
- Bthere exists a prime integer which divides $P(n)$.
- C$P(n)$ is divisible by $5$ for all $n \in N$.
- D$P(n)$ is divisible by $3$ for all $n \in N$.
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Similar Questions
Prove the statement by the Principle of Mathematical Induction: $\sqrt{n} < \frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + \ldots + \frac{1}{\sqrt{n}}$ for all natural numbers $n \geq 2$.
Difficult
View SolutionUse the Principle of Mathematical Induction to show that for a sequence $b_{0}, b_{1}, b_{2}, \ldots$ defined by $b_{0}=5$ and $b_{k}=4+b_{k-1}$ for all natural numbers $k$,the general term is $b_{n}=5+4n$ for all natural numbers $n$.
Difficult
View SolutionProve the following by using the principle of mathematical induction for all $n \in N$:
$1+3+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$
$1+3+3^{2}+\ldots+3^{n-1}=\frac{3^{n}-1}{2}$
Medium
View SolutionProve the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
$1 \cdot 3 + 2 \cdot 3^{2} + 3 \cdot 3^{3} + \ldots + n \cdot 3^{n} = \frac{(2n - 1) 3^{n+1} + 3}{4}$
Difficult
View SolutionUse the Principle of Mathematical Induction to prove that the number of subsets of a set containing $n$ distinct elements is $2^{n}$,for all $n \in N$.
Medium
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