# Prove the following by using the principle of mathematical induction for all $n\in \mathbb{N}$ :Q: 18        $1+2+3+...+n<\frac{1}{8}(2n+1)^2.$

Let the given statement be p(n) i.e.
$p(n):1+2+3+...+n<\frac{1}{8}(2n+1)^2.$
For n = 1  we have
$p(1):1<\frac{1}{8}(2(1)+1)^2= \frac{1}{8}(3)^2=\frac{9}{8}$   ,   which is true

For  n = k  we have

$p(k):1+2+3+...+k<\frac{1}{8}(2k+1)^2 \ \ \ \ \ \ \ \ \ -(i)$   ,        Let's assume that this statement is true

Now,
For  n = k + 1  we have
$p(k+1):1+2+3+...+k+1$                                                                                                                                                                                                                                                                              $=1+2+3+...+k+k+1$

$< \frac{1}{8}\left ( 2k+1 \right )^2+(k+1) \ \ \ \ \ \ \ \ \ (using \ (i))$
$< \frac{1}{8}\left ( (2k+1)^2+8(k+1) \right )$
$< \frac{1}{8}\left ( 4k^2+4k+1+8k+8 \right )$
$< \frac{1}{8}\left ( 4k^2+12k+9\right )$
$< \frac{1}{8}\left ( 2k+3\right )^2$
$< \frac{1}{8}\left ( 2(k+1)+1\right )^2$

Thus,  p(k+1)  is true whenever p(k) is true
Hence, by the principle of mathematical induction, statement p(n)  is true for all natural numbers n

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