Proving that $\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+\cdots+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}$ by induction

Question

I've proved the base case where $n=1$ and made the assumption that $n=k$ is true, but I'm stuck on the $n=k+1$ part. I just cannot seem to get the algebra to work in my favor.

Here is the original:

$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+…+\frac{1}{(n+3)(n+4)}=\frac{n}{4(n+4)}\qquad \forall\, n \in \mathbb N$

I get it to the following form and just run in circles:

$\displaystyle\frac{1}{4(5)}+\frac{1}{5(6)}+\frac{1}{6(7)}+...+\frac{1}{(k+3)(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$

Simplifying

$\displaystyle\frac{k}{4(k+4)} +\frac{1}{(k+4)(k+5)}=\frac{k+1}{4(k+1+4)}$

Is this in the correct form?

Answer 1

This should help: \begin{align} \sum_{i=1}^{k+1}\frac{1}{(i+3)(i+4)}&= \sum_{i=1}^k\frac{1}{(i+3)(i+4)}+\frac{1}{(k+4)(k+5)}\\[1em] &= \frac{k}{4(k+4)}+\frac{1}{(k+4)(k+5)}\\[1em] &= \frac{k(k+5)+4}{4(k+4)(k+5)}\\[1em] &= \frac{(k+1)(k+4)}{4(k+4)(k+5)}\\[1em] &= \frac{k+1}{4(k+5)}. \end{align} See if you can spot where the inductive hypothesis is used and where some basic algebraic manipulation plays an important role.

Answer 2

Let $S(n)$ be the statement: $\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(n+3)(n+4)}=\dfrac{n}{4\hspace{1 mm}(n+4)}$; $n\in\mathbb{N}$

Basis step: $S(1)$:

LHS: $\dfrac{1}{\big((1)+3\big)\big((1)+4\big)}=\dfrac{1}{4\times{5}}$

$\hspace{47.5 mm}=\dfrac{1}{20}$

RHS: $\dfrac{n}{4\hspace{1 mm}(n+4)}=\dfrac{(1)}{4\hspace{1 mm}\big((1)+4\big)}$

$\hspace{29 mm}=\dfrac{1}{4\times{5}}$

$\hspace{29 mm}=\dfrac{1}{20}$

$\hspace{80 mm}$ LHS $=$ RHS (verified.)

Inductive step:

Assume $S(k)$ is true, i.e. assume that

$\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(k+3)(k+4)}=\dfrac{k}{4\hspace{1 mm}(k+4)}$; $k\in\mathbb{N}$

$S(k+1)$: $\underline{\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(k+3)(k+4)}}+\dfrac{1}{\big((k+1)+3\big)\big((k+1)+4\big)}$

$\hspace{13 mm}=\dfrac{k}{4\hspace{1 mm}(k+4)}+\dfrac{1}{(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{k\hspace{1 mm}(k+4)(k+5)+4\hspace{1 mm}(k+4)}{4\hspace{1 mm}(k+4)(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{(k+4)\big(k\hspace{1 mm}(k+5)+4\big)}{4\hspace{1 mm}(k+4)^{2}(k+5)}$

$\hspace{13 mm}=\dfrac{k^{2}+5k+4}{4\hspace{1 mm}(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{k^{2}+k+4k+4}{4\hspace{1 mm}(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{k\hspace{1 mm}(k+1)+4\hspace{1 mm}(k+1)}{4\hspace{1 mm}(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{(k+1)(k+4)}{4\hspace{1 mm}(k+4)(k+5)}$

$\hspace{13 mm}=\dfrac{k+1}{4\hspace{1 mm}(k+5)}$

So, $S(k+1)$ is true whenever $S(k)$ is true.

Therefore, $\dfrac{1}{4(5)}+\dfrac{1}{5(6)}+\dfrac{1}{6(7)}+\cdots+\dfrac{1}{(n+3)(n+4)}=\dfrac{n}{4\hspace{1 mm}(n+4)}$; $n\in\mathbb{N}$.