Question

Prove the following by using the principle of mathematical induction for all $n\in N$
$\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Answer 1

Step $\left(1\right):$ Assume given statement
Let the given statement be $P\left(n\right)$, i.e.
$P\left(n\right):\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3n-1)(3n+2)}=\frac{n}{(6n+4)}$

Step $\left(2\right):$ Checking statement $P\left(n\right)$ for $n=1$
Put $n=1$ in $P\left(n\right)$, we get
$P\left(1\right):\frac{1}{2\cdot 5}=\frac{1}{6\cdot 1+4}$
$\Rightarrow \frac{1}{10}=\frac{1}{10}$
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 true for some natural number $K$ i.e.,
$P\left(k\right):\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3K-1)(3K+2)}$
$=\frac{K}{(6K+4)}\cdots \left(1\right)$

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
$\frac{1}{2\cdot 5}+\frac{1}{5\cdot 8}+\frac{1}{8\cdot 11}+\cdots +\frac{1}{(3K-1)(3K+2)}+\frac{1}{(3K+2)(3K+5)}$
$=\frac{K}{(6K+4)}+\frac{1}{(3K+2)(3K+5)}$ (using $\left(1\right)$)
$=\frac{K}{2(3K+2)}+\frac{1}{(3K+2)(3K+5)}$
$=\frac{1}{(3K+2)}(\frac{K}{2}+\frac{1}{3K+5})$
$=\frac{1}{(3K+2)}\left(\frac{3K^2+5K+2}{2(3K+5)}\right)$
$=\frac{1}{(3K+2)}\left(\frac{3K^2+2K+3K+2}{2(3K+5)}\right)$
$=\frac{1}{(3K+2)}\left(\frac{(3K+2)(K+1)}{2(3K+5)}\right)$
$=\frac{K+1}{2(3K+5)}$
$=\frac{K+1}{6K+10}$
$=\frac{(K+1)}{6(K+1)+4}$
Thus, $P(K+1)$ is the true whenever $P\left(K\right)$ is true.
Final answer :
Hence, from the principle of mathematical induction, the statement $P\left(n\right)$ is true for all $n\in N$.