Question

$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}.$

Answer 1

Let p(n)=$\frac{1}{1.4}+\frac{1}{4.7}+\frac{7}{7.10}+.....+\frac{1}{(3n-2)(3n+1)}=\frac{n}{3n+1}$

For n= 1

L.H.S$\frac{1}{1.4}=\frac{1}{4}$

R.H.S$\frac{1}{3\left(1\right)+1}=\frac{1}{4}$

$\therefore$ L.H.S=R.H.S

Thus p(n)is sum for n = 1

Assume that for p(k) is true.

$\therefore \frac{1}{1.4}+\frac{1}{1.4}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}=\frac{k}{3k+1}----\left(1\right)$

now,$\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{(3k-2)(3k+1)}+\frac{1}{\left(3\right(k+1)-2)\left(3\right(k+1)+1)}$

$=\frac{k}{3k+1}+\frac{1}{(3k+3-2)(3k+3+1)}$

$=\frac{k}{3k+1}+\frac{1}{(3k+1)(3k+4)}$

$=\frac{k(3k+4)+1}{(3k+1)(3k+4)}$

$=\frac{3k^2+4k+1}{(3k+1)(3k+4)}$

$=\frac{3k^2+3k+k+1}{(3k+1)(3k+4)}=\frac{3k(k+1)+(k+1)}{(3k+1)(3k+4)}$

$=\frac{(3k+1)(k+1)}{(3k+1)(3k+4)}$

$=\frac{k+1}{3k+4}$

$\therefore \frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{\left(3\right(k+1)-2)\left(3\right(k+1)+1)}-\frac{k+1}{3k+3+1}$

$=\frac{k+1}{3(k+1)+1}$

$\therefore$ p(k+1) is true whenever p(k) is true.

Hence by principle of mathematical Induction , p(n) is true

For n, where n is a natural number