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

Let $P(n)$ be the statement $1+2+\ldots+n<\dfrac18(2n+1)^2$.

Base case: For $n=1$,

Since $1<1.125$, $P(1)$ is true.

Inductive step: Assume $P(k)$ is true for some $k\ge1$:

We must show $1+2+\ldots+k+(k+1)<\dfrac18(2k+3)^2$.

Adding $(k+1)$ to both sides of the hypothesis:

Now compute $\dfrac18(2k+3)^2-\dfrac18(2k+1)^2$ using the difference of squares:

So …

Inequality induction. Add $(k+1)$ to the induction hypothesis, then show algebraically (via difference of squares) that $\frac18(2k+1)^2+(k+1)$ equals exactly $\frac …

• Trying to bound $(k+1)$ crudely instead of computing the exact identity $\frac18(2k+3)^2-\frac18(2k+1)^2=k+1$, which is what makes the step airtight.
• Sign error in the difference-of-squares expansion $(2k+3)^2-(2k+1)^2$. …