Prove the following by using the principle of mathematical induction for all n ∈ N:

Solution:

Let the given statement be P(n), i.e.,

$\mathrm{P}(n): 1+2+3+\ldots+n<\frac{1}{8}(2 n+1)^{2}$

It can be noted that $\mathrm{P}(n)$ is true for $n=1$ since $1<\frac{1}{8}(2.1+1)^{2}=\frac{9}{8}$.

Let P(k) be true for some positive integer k, i.e.,

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

We shall now prove that P(k + 1) is true whenever P(k) is true.

Consider

$(1+2+\ldots+k)+(k+1)<\frac{1}{8}(2 k+1)^{2}+(k+1) \quad[$ Using $(1)]$

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

$<\frac{1}{8}\left\{4 k^{2}+4 k+1+8 k+8\right\}$

$<\frac{1}{8}\left\{4 k^{2}+12 k+9\right\}$

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

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

Hence, $(1+2+3+\ldots+k)+(k+1)<\frac{1}{8}(2 k+1)^{2}+(k+1)$

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 i.e., n.