<s>[INST] How to prove that a statement is false using principle of mathematical induction? I came across this question while solving some problems based on principle of mathematical induction , 
$P(n) : 1+2+3+....+n < \frac{(n+2)^2}{8}, n\in\mathbb{N}$, is true for 
$(A) \, n\geq1\\
(B) \, n\geq2\\
(C) \text{ all } n\\
(D) \text{ none of these.}$ 
At first I kept some random values of $n$ in the inequality and found that it was true only for $n=1$ and not for any other natural value of $n$. Then I tried to prove the same or at least the fact that it is not true for all natural $n$ , using principle of mathematical induction. For this I first proved that $P(n)$ is true for $n=1,$ then I assumed that $P(n)$ is true for any natural number $n=k,$ i.e. 
$P(k) : 1+2+3+\dots+k < \frac{(k+2)^2}{8}$ is a true statement , then I tried to prove that $P(k+1)$ is a true statement using $P(k)$ i.e. $P(k+1) : 1+2+3+...+k+k+1 < \frac{(k+1+2)^2}{8} =\frac{(k+3)^2}{8}$ should be true . So $P(k) : 
1+2+3+....+k < \frac{(k+2)^2}{8},$ 
adding $k+1$ to both sides
$\begin{align}&\Rightarrow 1+2+...+k+k+1 
< \frac{(k+2)^2}{8} + k+1\\
&\Rightarrow 1+2+...+k+k+1 < \frac{k^2+4+4k+8k+8}{8} = \frac{k^2+12k+12}{8}\end{align}$
Now $\frac{k^2+12k+12}{8} = \frac{k^2+6k+9}{8}+ \frac{6k+3}{8}= \frac{(k+3)^2}{8} + \frac{6k+3}{8} \Rightarrow 
\frac{k^2+12k+12}{8}> \frac{(k+3)^2}{8}.$ So finally we have 
$1+2+...+k+k+1 < \frac{k^2+12k+12}{8}$ and 
$\frac{(k+3)^2}{8} < \frac{k^2+12k+12}{8},$ So from the above two inequalities we cannot prove that $P(k+1)$ is true but we also cannot prove it false , so what should we conclude from this? Also , if somehow we prove that $P(k+1)$ is false , is it not possible that the truth of $P(k)\Rightarrow$ truth of $P(k+2)$ and hence by principle of mathematical induction , $P(n)$ is true for alternate consecutive integers starting from $1$? Please help me.
 [/INST] Just prove by induction that $\sum_{j=1}^nj\ge\frac18(n+2)^2$ for all $n\ge2$, using $k+1\ge(2k+5)/8$ (which is equivalent to $6k\ge3$).
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