<s>[INST] mathematical induction proof aren't solving i am trying to solve for a mathematical induction for a while now but i seem to be getting the incorrect answer
$$
    \frac{1}{10}+\frac{1}{40}\:+\frac{1}{88}+_...+\frac{1}{\left(3n-1\right)\left(3n+2\right)}=\frac{n}{6n+4}
$$
Equation
i am unable to solve the left hand side to the right hand side i have tried multiple ways but I do not seem to be able to get it to be equal can some one please help
i am able to go up till this step
$$
\frac{k}{6k+4}+\frac{1}{\left(3\left(k+1\right)-1\right)\left(3\left(k+1\right)+2\right)}=\frac{k+1}{6\left(k+1\right)+4}
$$
i think am doing somthing incorrect after this
i tried following this youtube video
https://www.youtube.com/watch?v=twA6vZgX_U4
but it somehow doesnt work for this question
 [/INST] HINT: Write the $n^{\text{th}}$ term.
$$T_n = \frac{1}{(3n-1)(3n+2)} = \frac{1}{3}\frac{(3n+2) - (3n-1)}{(3n-1)(3n+2)} = \frac{1}{3}\frac{1}{3n-1} - \frac{1}{3}\frac{1}{3n+2}$$
Can you identify what happens on adding successive terms?

Using Mathematical Induction, first of all, verify the base case of $n=1$ which you can do. After that, assume that the given formula for some is true for $n$ terms i.e.
$$S_n = \frac{1}{10}+ \frac{1}{40} + \cdots + \frac{1}{(3n-1)(3n+2)} = \frac{n}{6n+4}$$
Now, for the sum till $n+1$ terms i.e.
$$S_{n+1} = \frac{1}{10}+ \frac{1}{40} + \cdots + \frac{1}{(3n-1)(3n+2)} + \frac{1}{(3n+2)(3n+5)}$$
you have
$$S_{n+1} = S_n + \frac{1}{(3n+2)(3n+5)} = \frac{n}{6n+4} + \frac{1}{(3n+2)(3n+5)}$$
This gives you
$$S_{n+1} = \frac{3n^2+5n+2}{(6n+4)(3n+5)} = \frac{(n+1)(3n+2)}{(6n+4)(3n+5)} = \frac{n+1}{6n+10} = \frac{n+1}{6(n+1) + 4}$$
Here, the RHS is the exactly what you get when you put $n+1$ in the proposed sum expression.
</s>