Here are the first four terms of a sequence. 2, $\frac{5}{3}$, $\frac{10}{5}$, $\frac{17}{7}$

1. Extend the numerator and denominator patterns to find the fifth term.

Write $2$ as $\dfrac{2}{1}$.

The terms are
$\frac{2}{1},\ \frac{5}{3},\ \frac{10}{5},\ \frac{17}{7}.$

Numerators: $2, 5, 10, 17$ increase by $3, 5, 7$, so the next increase is $9$.
$17+9=26$

Denominators: $1, 3, 5, 7$ increase by $2$, so the next denominator is
$7+2=9.$

Hence the fifth term is
$\frac{26}{9}.$

1. Identify the pattern in the numerators.

Write the terms as
$\frac{2}{1},\ \frac{5}{3},\ \frac{10}{5},\ \frac{17}{7}.$
The numerators are
$2,\ 5,\ 10,\ 17.$
These follow the pattern
$1^2+1,\ 2^2+1,\ 3^2+1,\ 4^2+1.$
So the numerator of the $n$th term is $n^2+1$.

2. Identify the denominator pattern and form the nth term.

The denominators are
$1,\ 3,\ 5,\ 7.$
These are consecutive odd numbers, so the denominator of the $n$th term is
$2n-1.$
Therefore,
$T_n=\frac{n^2+1}{2n-1}.$

1. Use the expression for $T_n$ to calculate $T_{25}-T_{24}$.

From part (b),
$T_n=\frac{n^2+1}{2n-1}.$
So
$T_{25}-T_{24}=\frac{25^2+1}{2(25)-1}-\frac{24^2+1}{2(24)-1}$
$=\frac{626}{49}-\frac{577}{47}$
$=\frac{626(47)-577(49)}{49(47)}$
$=\frac{29422-28273}{2303}=\frac{1149}{2303}.$