For all $n \in N$,$2^{2n+1} + 3^{2n+1}$ is divisible by

Let $P(n)$ denote the statement that $n^2 + n$ is odd. It is seen that $P(n) \Rightarrow P(n + 1)$. $P(n)$ is true for all:

Prove the following by using the principle of mathematical induction for all $n \in N$:
$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right) \dots \left(1+\frac{2n+1}{n^{2}}\right)=(n+1)^{2}$

Prove the following by using the principle of mathematical induction for all $n \in N$:
$1+2+3+\ldots+n < \frac{1}{8}(2n+1)^{2}$

Let $P(n) : 3^n < n!$ for $n \in N$ be true for $n \geq \lambda$. Find the smallest value of $\lambda$.

Using 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: