Question

Let by mathematical induction, for any natural numbers n,
$\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+.....+\frac{1}{(3n-1)(3n+2)}=\frac{n}{an+b}$. Find $a+b$

Answer 1

Solution: $\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+--+\frac{1}{(3n-1)(3n+2)}=\frac{n}{(an+b)}$

Let $p\left(n\right)=\frac{1}{2.5}+\frac{1}{5.8}+\frac{1}{8.11}+---+\frac{1}{(3n-1)(3n+2)}=\frac{n}{an+b}$

Consider $a=6,b=4$ form $=1$

$\therefore LHS=\frac{1}{2.5}=\frac{1}{10}$

$RHS=\frac{1}{\left[6\right(1)+4]}=\frac{1}{10}$

Hence, LHS = RHS

$\therefore a+b=6+4=10$