一个数列的第 $n$ 项为 $T_n = \frac{6n-5}{3n}$。 利用该公式求 $T_7$，并将答案表示为假分数。 解释为什么 $\tfrac{64}{33}$ 不是该数列中的一项。 证明 $\tfrac{1}{3} \le T_n < 2$。 另一个数列的前四项如下所示。 $T_1 = 4 = 2 \times 3 - 2$ $T_2 = 10 = 3 \times 4 - 2$ $T_3 = 18 = 4 \times 5 - 2$ $T_4 = 28 = 5 \times 6 - 2$ 求 $T_{10}$。 证明 $T_n = n^2 + 3n$。 已知 $T_k = 208$，利用 (b)(ii) 求 $k$ 的值。

1. Substitute $n=7$ into the formula and simplify.

$T_n=\dfrac{6n-5}{3n}$

$T_7=\dfrac{6(7)-5}{3(7)}=\dfrac{42-5}{21}=\dfrac{37}{21}$

1. Assume the given fraction is a term and solve for $n$.

If $\dfrac{64}{33}$ is a term, then
$\frac{6n-5}{3n}=\frac{64}{33}.$
Cross-multiplying,
$33(6n-5)=64(3n)$
$198n-165=192n$
$6n=165$
$n=27.5.$

2. Explain why this value of $n$ is invalid for a sequence term.

For a sequence, $n$ must be a positive integer. Since $27.5$ is not an integer, $\dfrac{64}{33}$ is not a term in the sequence.

1. Rewrite $T_n$ to make comparison with $2$ easier.

$T_n=\frac{6n-5}{3n}=\frac{6n}{3n}-\frac{5}{3n}=2-\frac{5}{3n}$

2. Use $n\geq 1$ to establish the lower and upper bounds for $T_n$.

Since $n$ is a positive integer, $n\geq 1$.

So
$0<\frac{5}{3n}\leq \frac{5}{3}.$

Therefore
$2-\frac{5}{3}\leq 2-\frac{5}{3n}<2,$
$\frac{1}{3}\leq T_n<2.$

1. Substitute the term number into the observed pattern and calculate $T_{10}$.

From the pattern,
$T_{10}=11\times 12-2=132-2=130.$

1. Identify the general product pattern from the given terms.

From the pattern,
$T_1=2\times3-2,$
$T_2=3\times4-2,$
$T_3=4\times5-2,$
$T_4=5\times6-2.$
So for the $n$th term,
$T_n=(n+1)(n+2)-2.$

2. Expand and simplify the expression to obtain the required form.

$T_n=(n+1)(n+2)-2$
$=n^2+3n+2-2$
$=n^2+3n$

1. Substitute $T_k=208$ into the formula from (b)(ii).

From (b)(ii),
$T_n=n^2+3n$
Given $T_k=208$,
$k^2+3k=208$

2. Rearrange and solve the quadratic equation by factorisation.

$k^2+3k-208=0$
$(k+16)(k-13)=0$
So
$k=-16 \quad \text{or} \quad k=13$

3. Choose the valid value of $k$ as a term number.

Since $k$ is a term number, $k$ must be a positive integer.

Hence
$k=13$