Question

Prove the following by using the principle of mathematical induction for all $n\in N.$
$1+2+3+4+\cdots +n<\frac{1}{8}(2n+1{)}^2$.

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$ i.e.,
$P\left(n\right):1+2+3+4+...+n<\frac{1}{8}(2n+1{)}^2$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):1<\frac{1}{8}(2\cdot 1+1{)}^2$
$\Rightarrow 1<\frac{1}{8}\cdot (3{)}^2$
$\Rightarrow 1<\frac{9}{8}$
Thus $P\left(n\right)$ is true for $n=1$

Step $\left(3\right):$ $P\left(n\right)$ for $n=K$.
Put $n=K$ in $P\left(n\right)$ and assume this is true for some natural number $K$ i.e.,
$P\left(k\right):1+2+3+4+\cdots +K<\frac{1}{8}(2K+1{)}^2\cdots (1)$

Step $\left(4\right):$ Checking statement $P\left(n\right)$ for $n=K+1$
Now, we shall prove that $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Now, we have
$(1+2+3+\cdots +K)+(K+1)$
$<\frac{1}{8}(2K+1{)}^2+(K+1)$ (using$\left(1\right)$)
$<\frac{1}{8}\left[\right(2K+1{)}^2+8(K+1)]$
$<\frac{1}{8}[4K^2+1+4K+8K+8]$
$<\frac{1}{8}[4K^2+12K+9]$
$<\frac{1}{8}(2K+3{)}^2$
$<\frac{1}{8}\left[2\right(K+1)+1{]}^2$
Hence, $1+2+3+\cdots +K+K+1<\frac{1}{8}\left\{2\right(K+1)+1{\}}^2$
Thus, $P(K+1)$ is true whenever $P\left(K\right)$ is true.
Final answer :
Therefore, by the principle of mathematical induction, statement $P\left(n\right)$ is true for all $n\in N$.