Question

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

Answer 1

Let $P\left(n\right)=1+2+3+\cdots +n<\frac{1}{8}(2n+1{)}^2.$
For n = 1
$P\left(1\right)=1<\frac{1}{8}(2\times 1+1{)}^2\Rightarrow 1<\frac{9}{8}\therefore P(1)\text{is true}\text{Let P (n) be true for n = k}\therefore P(k)=1+2+3+\cdots +K<\frac{1}{8}(2k+1{)}^2\text{For n = k + 1}P(k+1)=1+2+3+\cdots +k+(k+1)<\frac{1}{8}(2k+3{)}^2\text{From (i) we have}1+2+3+\cdots +k<\frac{1}{8}(2k+1{)}^2\text{Addingt (k + 1) on both sides, we get}1+2+3+\cdots +k+(k+1)<\frac{1}{8}(2k+1)62+(k+1)1+2+3+\cdots +k+(k+1)<\frac{1}{8}[4k^2+4k+1+8k+8]1+2+3+\cdots +k+(k+1)<\frac{1}{8}[4k62+12k+9]1+2+3+\cdots +k+(k+1)<\frac{1}{8}(2k+3{)}^2\frac{k+1}{2}+[2\times 1+(k+1-1)1]<\frac{1}{8}(2k+3{)}^2\Rightarrow \frac{(k+1)(k+2)}{2}<\frac{1}{8}(2k+3{)}^2\Rightarrow 4(k+1\left)\right(k+2)<4k62+9+6k\Rightarrow 4(k^2+3k+2)<4k^2+9+6k\Rightarrow 4k^2+12k+8<4k^2+9+6k\Rightarrow 8<9\therefore \text{P (k + 1) is true}$
Thus P (k ) is true $\Rightarrow$ P (k + 1) is true
hence by principle of mathematical induction,