Q14 (e)

Let $n=1$.

$\text{LHS}=u_1=1$

$\text{RHS}=\frac{1}{2(1)-1}=\frac11=1$

True for $n=1$.

Assume true for $n=k$, where $k$ is a positive integer.

$u_k=\frac{k}{2k-1}$

Prove true for $n=k+1$ if true for $n=k$.

Aim: Prove that
$u_{k+1}=\frac{k+1}{2(k+1)-1}$
$u_{k+1}=\frac{k+1}{2k+1}$

$\text{LHS}=u_{n+1}$
$=\frac{3u_k-1}{4u_k-1}$
$=\frac{3\left(\frac{k}{2k-1}\right)-1}{4\left(\frac{k}{2k-1}\right)-1}$
$=\frac{3k-(2k-1)}{4k-(2k-1)}$
$=\frac{k+1}{2k+1}$
$=\text{RHS}$

By Mathematical induction,
$u_n=\frac{n}{2n-1}$
is true for all integers, $n\ge1$.