Exercises: General Term of Sequences and Series

Solution: Alright, let’s break this down. First, let’s look at the differences between consecutive terms:

• $5 – 2 = 3$
• $10 – 5 = 5$
• $17 – 10 = 7$
• $26 – 17 = 9$

The first differences are $3, 5, 7, 9, \dots$. This isn’t constant, so it’s not arithmetic. Let’s look at the second differences:

• $5 – 3 = 2$
• $7 – 5 = 2$
• $9 – 7 = 2$

Aha! The second differences are constant, which tells us this is a quadratic sequence of the form $T_n = an^2 + bn + c$.

We know that $2a = (\text{second difference})$, so $2a = 2 \Rightarrow a = 1$.

Now, let’s use the first term and the formula:

• For $n=1$, $T_1 = a(1)^2 + b(1) + c = a+b+c = 2$.
• For $n=2$, $T_2 = a(2)^2 + b(2) + c = 4a+2b+c = 5$.
• For $n=3$, $T_3 = a(3)^2 + b(3) + c = 9a+3b+c = 10$.

Since $a=1$, we can substitute that in:

• $1+b+c = 2 \Rightarrow b+c = 1$ (Equation 1)
• $4+2b+c = 5 \Rightarrow 2b+c = 1$ (Equation 2)

Subtract Equation 1 from Equation 2:

$(2b+c) – (b+c) = 1 – 1 \Rightarrow b = 0$.

Substitute $b=0$ back into Equation 1:

$0+c = 1 \Rightarrow c = 1$.

So, the general term is $T_n = n^2 + 0n + 1$, which simplifies to:

$$T_n = n^2 + 1$$

Let’s do a quick check: $T_1 = 1^2+1 = 2$, $T_2 = 2^2+1 = 5$, $T_3 = 3^2+1 = 10$. Looks good!

Solution: Don’t let the fractions intimidate you! Let’s look at the numerators and denominators separately.

Numerator: $1, 2, 3, 4, 5, \dots$

This is pretty straightforward, the numerator for the $n$-th term is just $n$. So, Numerator $= n$.

Denominator: $2, 5, 10, 17, 26, \dots$

Hold on, this looks familiar! If you recall Question 1, this is exactly the sequence $n^2 + 1$. Let’s quickly verify:

• For $n=1$, $1^2+1 = 2$
• For $n=2$, $2^2+1 = 5$
• For $n=3$, $3^2+1 = 10$
• For $n=4$, $4^2+1 = 17$
• For $n=5$, $5^2+1 = 26$

Yep, that’s it! So, Denominator $= n^2 + 1$.

Putting it all together, the general term for the sequence is:

$$T_n = \frac{n}{n^2+1}$$

Solution: Let’s get into the rhythm. Again, we’ll start by finding the differences:

• $6 – 3 = 3$
• $11 – 6 = 5$
• $18 – 11 = 7$
• $27 – 18 = 9$

The first differences are $3, 5, 7, 9, \dots$. Not constant yet.

Now for the second differences:

• $5 – 3 = 2$
• $7 – 5 = 2$
• $9 – 7 = 2$

Boom! Constant second differences means it’s a quadratic sequence, $T_n = an^2 + bn + c$.

From the second differences, $2a = 2 \Rightarrow a = 1$.

Let’s substitute $a=1$ into the first few terms:

• $T_1 = 1^2 + b(1) + c = 1+b+c = 3$
• $T_2 = 1(2)^2 + b(2) + c = 4+2b+c = 6$

From $1+b+c = 3 \Rightarrow b+c = 2$ (Equation 1)

From $4+2b+c = 6 \Rightarrow 2b+c = 2$ (Equation 2)

Subtract Equation 1 from Equation 2:

$(2b+c) – (b+c) = 2 – 2 \Rightarrow b = 0$.

Substitute $b=0$ back into Equation 1:

$0+c = 2 \Rightarrow c = 2$.

So, the general term is $T_n = n^2 + 0n + 2$, which simplifies to:

$$T_n = n^2 + 2$$

Let’s check: $T_1 = 1^2+2 = 3$, $T_2 = 2^2+2 = 6$, $T_3 = 3^2+2 = 11$. All good!

Solution: Okay, this sequence has a few things going on: fractions and alternating signs. Let’s tackle them one by one.

Signs: The terms alternate between negative and positive (starting with negative). This usually means a factor of $(-1)^n$ or $(-1)^{n+1}$.

• For $n=1$, we have a negative term. If we use $(-1)^n$, then $(-1)^1 = -1$, which matches.
• For $n=2$, we have a positive term. If we use $(-1)^n$, then $(-1)^2 = 1$, which matches.

So, the sign factor is $(-1)^n$.

Numerators: Every numerator is $1$. Easy peasy!

Denominators: $1, 2, 6, 24, 120, \dots$

Do these numbers ring a bell? Let’s write them out and see if we can find a pattern:

• $1 = 1!$ (factorial of 1)
• $2 = 2!$ (factorial of 2)
• $6 = 3!$ (factorial of 3)
• $24 = 4!$ (factorial of 4)
• $120 = 5!$ (factorial of 5)

Yep, the denominator for the $n$-th term is $n!$ (n factorial).

Putting all the pieces together:

$$T_n = (-1)^n \frac{1}{n!}$$

Let’s quickly test it:

• $T_1 = (-1)^1 \frac{1}{1!} = -1 \cdot \frac{1}{1} = -1$
• $T_2 = (-1)^2 \frac{1}{2!} = 1 \cdot \frac{1}{2} = \frac{1}{2}$
• $T_3 = (-1)^3 \frac{1}{3!} = -1 \cdot \frac{1}{6} = -\frac{1}{6}$

Looks perfect! Nicely done if you spotted the factorials!

Solution: Let’s follow our usual strategy: find the differences. If it’s a higher-order polynomial, we might need to go a few rounds.

First differences:

• $7 – 0 = 7$
• $26 – 7 = 19$
• $63 – 26 = 37$
• $124 – 63 = 61$

Sequence of first differences: $7, 19, 37, 61, \dots$ (Not constant)

Second differences:

• $19 – 7 = 12$
• $37 – 19 = 18$
• $61 – 37 = 24$

Sequence of second differences: $12, 18, 24, \dots$ (Still not constant, but looks arithmetic!)

Third differences:

• $18 – 12 = 6$
• $24 – 18 = 6$

Yes! The third differences are constant ($6$). This means our sequence is a cubic polynomial of the form $T_n = an^3 + bn^2 + cn + d$.

For a cubic sequence, we know that $6a = (\text{third difference})$. So, $6a = 6 \Rightarrow a = 1$.

Now, let’s use a systematic approach or look for a simpler pattern. Since $a=1$, maybe $T_n$ is close to $n^3$. Let’s compare $n^3$ to $T_n$:

• $n=1$: $1^3 = 1$. $T_1 = 0$. Difference: $0 – 1 = -1$.
• $n=2$: $2^3 = 8$. $T_2 = 7$. Difference: $7 – 8 = -1$.
• $n=3$: $3^3 = 27$. $T_3 = 26$. Difference: $26 – 27 = -1$.
• $n=4$: $4^3 = 64$. $T_4 = 63$. Difference: $63 – 64 = -1$.
• $n=5$: $5^3 = 125$. $T_5 = 124$. Difference: $124 – 125 = -1$.

It looks like each term $T_n$ is exactly $n^3 – 1$!

So, the general term is:

$$T_n = n^3 – 1$$