Question

$1+2+3+..............n<\frac{1}{8}(2n+1{)}^2$

Answer 1

TO PROVE :

$1+2+3+....n<\frac{1}{8}{(2n+1)}^2$

PROOF:

$P\left(n\right):1+2+3+....n<\frac{1}{8}{(2n+1)}^2$

$P\left(1\right):1<\frac{1}{8}{(2+1)}^2$

$\Rightarrow 1<\frac{1}{8}\times 9$

$\Rightarrow 1<\frac{9}{8}$ which is true.

$\therefore P\left(1\right)$ is true.

Let $P\left(m\right)$ is true.

Then, $P\left(m\right):1+2+3+....m<\frac{1}{8}{(2m+1)}^2$

Now, we prove it for $P(m+1)$

Then,

$P\left(m\right):1+2+3+....m+(m+1)$

$<\frac{1}{8}{(2m+1)}^2+(m+1)$

$<\frac{1}{8}({4m}^2+1+4m)+(m+1)$

$<\frac{{4m}^2+1+4m+8m+8}{8}$

$<\frac{{4m}^2+12m+9}{8}$

$<\frac{1}{8}{(2m+3)}^2$

$<\frac{1}{8}{[2(m+1)+1]}^2$ (Proved)

$\therefore P(m+1)$ is also true (proved).