<s>[INST] Beginner Proof by Induction: Is this correct? I am asked to prove $\frac{1}{1.4} + \frac{1}{4.7} + ... + \frac{1}{(3n-2)(3n+1)} = \frac{n}{3n+1}, n\geq 1$ by induction.
Can someone verify that I did this correctly. I am unsure about my Inductive Step.
Proof by induction
Inductive Hypothesis Let $P(k) = \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} = \frac{k}{3k+1}$
Base case $n = 1$
\begin{align*}
        P(1) &= \sum\limits_{i=1}^{1} \frac{1}{(3(1)-2)(3(1)+1)} = \frac{1}{3(1)+1} \\
        &= \frac{1}{(3-2)(3+1)} = \frac{1}{3+1} \\
        &= \frac{1}{(1)(4)} = \frac{1}{4} \\
        &= \frac{1}{4} = \frac{1}{4} \checkmark
    \end{align*}
Inductive Step
For $k \geq 1$, show that $P(k) \to P(k+1)$ is true.
Assume $P(k)$ is true, assume $\sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} = \frac{k}{3k+1}$ is true.
Prove $P(k+1)$ is true, examine $\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} = \frac{k+1}{3(k+1)+1}$.
\begin{align*}
    \sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \frac{k+1}{3(k+1)+1} \\
    \sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \frac{k}{3(k+1)} \\
    \sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} &= \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)}
\end{align*}

Can someone check that this is the correct way to prove the theorem by induction?
 [/INST] I don't understand what you're doing in the last couple of lines for $P(k + 1)$, for example between the third & second last, you change the RHS value but leave the LHS the same.
Instead, I would normally go from one side of the equation, expand or otherwise manipulate it as need be so can use the assumption that $P(k)$ is true, and then do various simplifications to show how it's the same as the other side. In general, I've found it's easiest to start with the summation. As such, here's how I would handle the inductive step.
$$\begin{equation}\begin{aligned}
\sum\limits_{i=1}^{k+1} \frac{1}{(3i-2)(3i+1)} & = \sum\limits_{i=1}^{k} \frac{1}{(3i-2)(3i+1)} + \frac{1}{(3(k+1)-2)(3(k+1)+1)} \\
& = \frac{k}{3k+1} + \frac{1}{(3k+1)(3k+4)} \\
& = \frac{k(3k+4) + 1}{(3k+1)(3k+4)} \\
& = \frac{3k^2 + 4k + 1}{(3k+1)(3k+4)} \\
& = \frac{(3k + 1)(k + 1}{(3k+1)(3k+4)} \\
& = \frac{k + 1}{3(k+1) + 1} \\
& = P(k+1)
\end{aligned}\end{equation}\tag{1}\label{eq1A}$$
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