(B) Given $P(n) = 3^{2n+1} + 2^{n+2}$.For $n = 1$, $P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$.Since $35 = 5 \times 7$, $P(1)$ is divisibl

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there exists a prime integer which divides $P(n)$.

(B) Given $P(n) = 3^{2n+1} + 2^{n+2}$.For $n = 1$, $P(1) = 3^{2(1)+1} + 2^{1+2} = 3^3 + 2^3 = 27 + 8 = 35$.Since $35 = 5 \times 7$, $P(1)$ is divisible by the prime integers $5$ and $7$.Thus, there exists at least one prime integer that divides $P(n)$ for $n = 1$.For any $n \in N$, $P(n) > 1$, and by the Fundamental Theorem of Arithmetic, every integer greater than $1$ has at least one prime divisor.Therefore, there exists a prime integer which divides $P(n)$.

Class 11 Mathematics Mathematical induction Mathematical induction

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Let $P(n) = 3^{2n+1} + 2^{n+2}$ where $n \in N$. Then

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A
$P(n)$ is not divisible by any prime integer.
B
there 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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Prove the statement by the Principle of Mathematical Induction: For any natural number $n$,$x^{n}-y^{n}$ is divisible by $x-y$,where $x$ and $y$ are any integers with $x \neq y$.

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Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

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Prove the following by using the principle of mathematical induction for all $n \in N$:
$1 \cdot 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \ldots + n \cdot 2^{n} = (n-1) 2^{n+1} + 2$

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Prove the following by using the principle of mathematical induction for all $n \in N$:
$\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$

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Prove the following by using the principle of mathematical induction for all $n \in N$ :
$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

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Let $P(n) = 3^{2n+1} + 2^{n+2}$ where $n \in N$. Then

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Prove the statement by the Principle of Mathematical Induction: For any natural number $n$,$x^{n}-y^{n}$ is divisible by $x-y$,where $x$ and $y$ are any integers with $x \neq y$.

Prove the following by using the principle of mathematical induction for all $n \in N$:$\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\ldots+\frac{1}{2^{n}}=1-\frac{1}{2^{n}}$

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Prove the following by using the principle of mathematical induction for all $n \in N$:$\frac{1}{2 \times 5} + \frac{1}{5 \times 8} + \frac{1}{8 \times 11} + \ldots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$

Prove the following by using the principle of mathematical induction for all $n \in N$ :$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$

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Prove the following by using the principle of mathematical induction for all $n \in N$:
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Prove the following by using the principle of mathematical induction for all $n \in N$:
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Prove the following by using the principle of mathematical induction for all $n \in N$:
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$\frac{1}{1 \cdot 2 \cdot 3} + \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{3 \cdot 4 \cdot 5} + \ldots + \frac{1}{n(n+1)(n+2)} = \frac{n(n+3)}{4(n+1)(n+2)}$