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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.

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  • $\begingroup$ For this one question, you might be overthinking it. P(1) is true. P(2) is false. So the answer is D. You could make a stronger statement about P by proving what J.G. suggests, but that's more than a basic reading of the question requires. $\endgroup$
    – The Chaz 2.0
    Commented Jan 2, 2020 at 17:22
  • $\begingroup$ The claim is only true for $n=1$ . Which option supports this claim ? $\endgroup$
    – The Demonix _ Hermit
    Commented Jan 2, 2020 at 17:22

1 Answer 1

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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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  • $\begingroup$ what about my second query , that , if by principle of mathematical induction we can't prove that for a given statement truth of P(k) does not imply truth of P(k+1) then is it not possible that truth of P(k) implies truth of P(k+2) ? $\endgroup$
    – Sameer Nilkhan
    Commented Jan 4, 2020 at 4:14
  • $\begingroup$ also can we conclude that whenever proving an inequality if truth of P(k) does not imply truth of P(k+1) then then the inequality with the sign of inequality reversed will surely be true ? $\endgroup$
    – Sameer Nilkhan
    Commented Jan 4, 2020 at 4:48
  • $\begingroup$ @Sameernilkhan You have to use an if-true-of-$k$-then-true-of-$k+1$ structure, which in this case works fine. But some creativity is possible. If you genuinely found a theorem of the form $P(k)\to P(k+2)$, the definition $Q(k):=P(k)\land P(k+1)$ would let you continue with an inductive argument. (It's also possible to show weak induction implies strong induction.) $\endgroup$
    – J.G.
    Commented Jan 4, 2020 at 8:05
  • $\begingroup$ please correct me if i am wrong for my second comment. $\endgroup$
    – Sameer Nilkhan
    Commented Jan 4, 2020 at 8:25

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