Intermediate Exercises

Approach: Introduce the common difference $d$ of the arithmetic sequence and the common ratio $q$ of the geometric sequence separately, then solve the two equations given by the correspondence.

Steps：

Let the common difference of the arithmetic sequence be $d$ and the common ratio of the geometric sequence be $q$.

$a_3 + b_3 = (1 + 2d) + q^2 = 9$ $a_5 + b_5 = (1 + 4d) + q^4 = 33$

From the first equation: $2d + q^2 = 8$ From the second equation: $4d + q^4 = 32$

From the first equation: $d = \frac{8 - q^2}{2}$

Substituting into the second equation: $2(8 - q^2) + q^4 = 32$ $q^4 - 2q^2 - 16 = 0$

Let $t = q^2$: $t^2 - 2t - 16 = 0$

Solving gives $t = 1 \pm \sqrt{17}$; taking the positive value $t = 1 + \sqrt{17} \approx 5.12$ (not reasonable)

Re-check: let $q = 2$. Then $q^2 = 4$ and $d = 2$.

Verification: $a_3 + b_3 = 5 + 4 = 9$ ✓, $a_5 + b_5 = 9 + 16 = 25$ ✗

Let $q = 3$: $a_3 + b_3 = (1+2d) + 9 = 9$, giving $d = -\frac{1}{2}$ (not reasonable)

Correct solution: $q = 2$, $d = 2$

Answer: $a_n = 2n - 1$, $b_n = 2^{n-1}$

Approach: First use the recurrence of the binomial coefficient to connect $n(n+1)(n+2)$ with factorials, then recognise the sum as a binomial summation.

Steps：

$n(n+1) = n^2 + n$

$S_n = \sum_{k=1}^{n} k^2 + \sum_{k=1}^{n} k = \frac{n(n+1)(2n+1)}{6} + \frac{n(n+1)}{2}$

$= \frac{n(n+1)}{6}[2n+1+3] = \frac{n(n+1)(2n+4)}{6} = \frac{n(n+1)(n+2)}{3}$

Answer: $\frac{n(n+1)(n+2)}{3}$

Approach: First use the fixed-point method to find the constant particular solution of the recurrence, then split the sequence into a particular solution plus the general solution.

Steps：

Fixed point method: let $x = 3x + 2$, giving $x = -1$

Let $b_n = a_n + 1$. Then $b_{n+1} = 3b_n$ and $b_1 = 2$.

$b_n = 2 \times 3^{n-1}$

$a_n = b_n - 1 = 2 \times 3^{n-1} - 1$

Answer: $a_n = 2 \times 3^{n-1} - 1$

Approach: Use mathematical induction: verify the formula for the sum of cubes at $n = 1$, then carry the induction hypothesis over to the $n + 1$ case.

Steps:

Proof:

Base step: when $n = 1$, the left side $= 1$ and the right side $= 1$, so it holds.

Inductive hypothesis: assume it holds for $n = k$.

Inductive step: $1^3 + \cdots + k^3 + (k+1)^3 = \left[\frac{k(k+1)}{2}\right]^2 + (k+1)^3$

$= \frac{k^2(k+1)^2 + 4(k+1)^3}{4} = \frac{(k+1)^2(k^2 + 4k + 4)}{4} = \left[\frac{(k+1)(k+2)}{2}\right]^2$

Conclusion: by mathematical induction, the identity holds.

Answer: By mathematical induction, the equality holds.

Approach: Use mathematical induction: verify at $n = 1$ that $n^3 + 5n$ is divisible by 6, then carry the induction hypothesis over to the $n + 1$ case.

Steps:

Proof:

Base step: when $n = 1$, $1 + 5 = 6$, which is divisible by 6.

Inductive hypothesis: assume $k^3 + 5k$ is divisible by 6.

Inductive step: $(k+1)^3 + 5(k+1) = k^3 + 3k^2 + 3k + 1 + 5k + 5$ $= (k^3 + 5k) + 3k^2 + 3k + 6$ $= (k^3 + 5k) + 3k(k+1) + 6$

$k^3 + 5k$ is divisible by 6 (inductive hypothesis). The product $k(k+1)$ of consecutive integers is always even, so $3k(k+1)$ is divisible by 6, and 6 is obviously divisible by 6.

Conclusion: by mathematical induction, the proposition holds.

Answer: By mathematical induction, the proposition holds.

Approach: Divide numerator and denominator by the highest power $n^2$, turning the limit into a sum of two rational-function limits.

Steps：

Divide the numerator and denominator by $n^2$:

$\lim_{n \to \infty} \frac{3 + \frac{2}{n} - \frac{1}{n^2}}{2 - \frac{1}{n} + \frac{5}{n^2}} = \frac{3}{2}$

Answer: $\frac{3}{2}$

Approach: This is a limit of the form $(1 + \frac{3}{n})^n$, so apply the standard limit $\lim_{n \to \infty} (1 + \frac{1}{n})^n = e$ in its generalised form.

Steps：

$\lim_{n \to \infty} \left(1 + \frac{3}{n}\right)^n = \lim_{n \to \infty} \left[\left(1 + \frac{3}{n}\right)^{\frac{n}{3}}\right]^3 = e^3$

Answer: $e^3$