2.3 MCQs-Sequences and Series

Sequences and Series

Basic Concepts and Definitions

1. A sequence is defined as:

1. A sum of numbers

2. A list of numbers arranged in a specific order

3. An unordered collection of numbers

4. The limit of a function

Show me the answer

Answer: 2. A list of numbers arranged in a specific order

Explanation: A sequence is an ordered list of numbers: $a_1, a_2, a_3, \ldots, a_n, \ldots$ Each number is called a term of the sequence. The position of each term is important, with $a_n$ representing the nth term. A sequence can be finite or infinite.

Example: $1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}, \ldots$ is an infinite sequence.

2. Which of the following represents an arithmetic sequence?

1. $2, 4, 8, 16, 32$

2. $3, 6, 9, 12, 15$

3. $1, 4, 9, 16, 25$

4. $1, 1, 2, 3, 5, 8$

Show me the answer

Answer: 2. $3, 6, 9, 12, 15$

Explanation: An arithmetic sequence has a constant difference between consecutive terms.

For the sequence $3, 6, 9, 12, 15$: $6 - 3 = 3$ $9 - 6 = 3$ $12 - 9 = 3$ $15 - 12 = 3$

The common difference is constant: $d = 3$.

General form: $a_n = a_1 + (n-1)d$

Option 1 is a geometric sequence, option 3 represents squares of natural numbers, and option 4 is the Fibonacci sequence.

3. Which of the following represents a geometric sequence?

1. $5, 10, 15, 20, 25$

2. $2, 4, 6, 8, 10$

3. $3, 9, 27, 81, 243$

4. $1, 3, 6, 10, 15$

Show me the answer

Answer: 3. $3, 9, 27, 81, 243$

Explanation: A geometric sequence has a constant ratio between consecutive terms.

For the sequence $3, 9, 27, 81, 243$: $\frac{9}{3} = 3$ $\frac{27}{9} = 3$ $\frac{81}{27} = 3$ $\frac{243}{81} = 3$

The common ratio is constant: $r = 3$.

General form: $a_n = a_1 \cdot r^{n-1}$

Options 1 and 2 are arithmetic sequences, option 4 represents triangular numbers.

4. The nth term of the sequence $2, 5, 8, 11, \ldots$ is:

1. $3n - 1$

2. $2n + 1$

3. $3n + 2$

4. $2n + 3$

Show me the answer

Answer: 1. $3n - 1$

Explanation: This is an arithmetic sequence with: First term: $a_1 = 2$ Common difference: $d = 3$

General formula for arithmetic sequence: $a_n = a_1 + (n-1)d$

Substitute values: $a_n = 2 + (n-1) \cdot 3$ $a_n = 2 + 3n - 3$ $a_n = 3n - 1$

Verification: $n=1: 3 \cdot 1 - 1 = 2$ $n=2: 3 \cdot 2 - 1 = 5$ $n=3: 3 \cdot 3 - 1 = 8$ $n=4: 3 \cdot 4 - 1 = 11$

Arithmetic Progressions (AP)

5. The sum of first n terms of an AP is given by:

1. $S_n = \frac{n}{2} [2a + (n-1)d]$

2. $S_n = \frac{a(r^n - 1)}{r - 1}$

3. $S_n = \frac{n(n+1)}{2}$

4. $S_n = \frac{a}{1 - r}$

Show me the answer

Answer: 1. $S_n = \frac{n}{2} [2a + (n-1)d]$

Explanation: For an AP with first term $a$ and common difference $d$, the sum of first n terms is:

$S_n = \frac{n}{2} [2a + (n-1)d]$

Derivation: Write the sum forwards and backwards, then add:

Alternative form: $S_n = \frac{n}{2} [\text{first term} + \text{last term}] = \frac{n}{2} (a + a_n)$

6. If the 5th term of an AP is 17 and the 9th term is 33, the common difference is:

1. $2$

2. $3$

3. $4$

4. $5$

Show me the answer

Answer: 3. $4$

Explanation: For an AP: $a_n = a + (n-1)d$

Given: $a_5 = a + 4d = 17 \quad \text{(1)}$ $a_9 = a + 8d = 33 \quad \text{(2)}$

Subtract equation (1) from equation (2): $(a + 8d) - (a + 4d) = 33 - 17$ $4d = 16$ $d = 4$

To find $a$: From equation (1): $a + 4 \cdot 4 = 17$ $a + 16 = 17$ $a = 1$

The AP is: $1, 5, 9, 13, 17, 21, 25, 29, 33, \ldots$

7. The arithmetic mean between 7 and 19 is:

1. $11$

2. $12$

3. $13$

4. $14$

Show me the answer

Answer: 3. $13$

Explanation: The arithmetic mean (AM) between two numbers $a$ and $b$ is:

$\text{AM} = \frac{a + b}{2}$

Substitute $a = 7$, $b = 19$: $\text{AM} = \frac{7 + 19}{2} = \frac{26}{2} = 13$

Alternatively, consider an AP of three terms: $7, x, 19$ The common difference must be constant: $x - 7 = 19 - x$ $2x = 26$ $x = 13$

The arithmetic mean is also called the average of two numbers.

Geometric Progressions (GP)

8. The sum of first n terms of a GP is given by:

1. $S_n = \frac{n}{2} [2a + (n-1)d]$

2. $S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{for } r \neq 1$

3. $S_n = \frac{a}{1 - r}$

4. $S_n = \frac{n(n+1)}{2}$

Show me the answer

Answer: 2. $S_n = \frac{a(1 - r^n)}{1 - r} \quad \text{for } r \neq 1$

Explanation: For a GP with first term $a$ and common ratio $r$:

For $r = 1$: $S_n = na$

Alternative form: $S_n = \frac{a(r^n - 1)}{r - 1} \quad \text{for } r \neq 1$

9. The geometric mean between 4 and 16 is:

1. $6$

2. $8$

3. $10$

4. $12$

Show me the answer

Answer: 2. $8$

Explanation: The geometric mean (GM) between two positive numbers $a$ and $b$ is:

$\text{GM} = \sqrt{ab}$

Substitute $a = 4$, $b = 16$: $\text{GM} = \sqrt{4 \times 16} = \sqrt{64} = 8$

Alternatively, consider a GP of three terms: $4, x, 16$ The common ratio must be constant: $\frac{x}{4} = \frac{16}{x}$ $x^2 = 64$ $x = 8$ (positive value)

For n numbers, $\text{GM} = (a_1 \times a_2 \times \cdots \times a_n)^{1/n}$

10. If the 3rd term of a GP is 12 and the 6th term is 96, the common ratio is:

1. $2$

2. $3$

3. $4$

4. $6$

Show me the answer

Answer: 1. $2$

Explanation: For a GP: $a_n = ar^{n-1}$

Given: $a_3 = ar^2 = 12 \quad \text{(1)}$ $a_6 = ar^5 = 96 \quad \text{(2)}$

Divide equation (2) by equation (1): $\frac{ar^5}{ar^2} = \frac{96}{12}$ $r^3 = 8$ $r = 2$ (since $2^3 = 8$)

To find $a$: From equation (1): $a \cdot 2^2 = 12$ $a \cdot 4 = 12$ $a = 3$

The GP is: $3, 6, 12, 24, 48, 96, \ldots$

Infinite Geometric Series

11. The sum to infinity of a geometric series exists when:

1. $|r| < 1$

2. $|r| > 1$

3. $r = 1$

4. $r = -1$

Show me the answer

Answer: 1. $|r| < 1$

Explanation: For an infinite geometric series: $S = a + ar + ar^2 + ar^3 + \cdots$

The sum to infinity exists when $|r| < 1$ and is given by: $S_{\infty} = \frac{a}{1 - r}$

Reason: As $n \to \infty$, $r^n \to 0$ when $|r| < 1$.

If $|r| \geq 1$, the series diverges (the sum goes to infinity or oscillates).

Example: $1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + \cdots = \frac{1}{1 - \frac{1}{2}} = 2$

12. The sum to infinity of the series $1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} + \cdots$ is:

1. $\frac{3}{2}$

2. $2$

3. $3$

4. $\infty$

Show me the answer

Answer: 1. $\frac{3}{2}$

Explanation: This is an infinite geometric series with: First term: $a = 1$ Common ratio: $r = \frac{1}{3}$

Since $|r| = \frac{1}{3} < 1$, the sum to infinity exists: $S_{\infty} = \frac{a}{1 - r} = \frac{1}{1 - \frac{1}{3}} = \frac{1}{\frac{2}{3}} = \frac{3}{2}$

Verification using partial sums: $S_1 = 1 = 1$ $S_2 = 1 + \frac{1}{3} = \frac{4}{3} \approx 1.333$ $S_3 = 1 + \frac{1}{3} + \frac{1}{9} = \frac{13}{9} \approx 1.444$ $S_4 = 1 + \frac{1}{3} + \frac{1}{9} + \frac{1}{27} = \frac{40}{27} \approx 1.481$ Approaching $1.5 = \frac{3}{2}$

Harmonic Progression (HP)

13. A sequence is said to be in harmonic progression if:

1. The reciprocals of its terms are in arithmetic progression

2. The terms have a constant difference

3. The terms have a constant ratio

4. The terms are reciprocals of natural numbers

Show me the answer

Answer: 1. The reciprocals of its terms are in arithmetic progression

Explanation: A sequence $h_1, h_2, h_3, \ldots$ is in harmonic progression if the sequence of reciprocals $\frac{1}{h_1}, \frac{1}{h_2}, \frac{1}{h_3}, \ldots$ is in arithmetic progression.

Example: $\frac{1}{2}, \frac{1}{4}, \frac{1}{6}, \frac{1}{8}, \ldots$ is in HP because the reciprocals $2, 4, 6, 8, \ldots$ are in AP.

The nth term of HP: $h_n = \frac{1}{a + (n-1)d}$ where $a = \frac{1}{h_1}$ and $d = \frac{1}{h_2} - \frac{1}{h_1}$.

Harmonic mean between a and b: $\text{HM} = \frac{2ab}{a+b}$

14. The harmonic mean between 2 and 8 is:

1. $3.2$

2. $4$

3. $5$

4. $6.4$

Show me the answer

Answer: 1. $3.2$

Explanation: The harmonic mean between two numbers $a$ and $b$ is:

$\text{HM} = \frac{2ab}{a+b}$

Substitute $a = 2$, $b = 8$: $\text{HM} = \frac{2 \times 2 \times 8}{2 + 8} = \frac{32}{10} = 3.2$

Verification: The HP with three terms would be: $2, 3.2, 8$ Reciprocals: $\frac{1}{2} = 0.5$, $\frac{1}{3.2} = 0.3125$, $\frac{1}{8} = 0.125$

Differences: $0.5 - 0.3125 = 0.1875$ $0.3125 - 0.125 = 0.1875$

The differences are constant, so the reciprocals are in AP, hence the original numbers are in HP.

Special Series

15. The sum of first n natural numbers is:

1. $\frac{n(n+1)}{2}$

2. $\frac{n(n-1)}{2}$

3. $n^2$

4. $\frac{n(n+1)(2n+1)}{6}$

Show me the answer

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

Explanation: The sum of first n natural numbers:

$1 + 2 + 3 + \cdots + n = \frac{n(n+1)}{2}$

Derivation (Gauss method): Let $S = 1 + 2 + 3 + \cdots + n$ Write backwards: $S = n + (n-1) + (n-2) + \cdots + 1$ Add the two equations:

Example: Sum of first 100 natural numbers = $\frac{100 \times 101}{2} = 5050$

16. The sum of squares of first n natural numbers is:

1. $\frac{n(n+1)}{2}$

2. $\frac{n(n+1)(2n+1)}{6}$

3. $\left[\frac{n(n+1)}{2}\right]^2$

4. $\frac{n(n+1)(n+2)}{6}$

Show me the answer

Answer: 2. $\frac{n(n+1)(2n+1)}{6}$

Explanation: The sum of squares of first n natural numbers:

$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$

Derivation can be done using mathematical induction or telescoping sums.

Verification: $n=1: 1^2 = 1, \quad \text{formula: } \frac{1 \times 2 \times 3}{6} = 1$ $n=2: 1^2+2^2 = 1+4=5, \quad \text{formula: } \frac{2 \times 3 \times 5}{6} = 5$ $n=3: 1^2+2^2+3^2 = 1+4+9=14, \quad \text{formula: } \frac{3 \times 4 \times 7}{6} = 14$

This formula is useful in calculus for Riemann sums and calculating areas under curves.

17. The sum of cubes of first n natural numbers is:

1. $\frac{n(n+1)(2n+1)}{6}$

2. $\left[\frac{n(n+1)}{2}\right]^2$

3. $\frac{n^2(n+1)^2}{4}$

4. Both 2 and 3

Show me the answer

Answer: 4. Both 2 and 3

Explanation: The sum of cubes of first n natural numbers:

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

Note that $\left[\frac{n(n+1)}{2}\right]^2 = \frac{n^2(n+1)^2}{4}$, so options 2 and 3 are equivalent.

Remarkable fact: $(1 + 2 + 3 + \cdots + n)^2 = 1^3 + 2^3 + 3^3 + \cdots + n^3$

Example for n=3: $(1+2+3)^2 = 6^2 = 36$ $1^3+2^3+3^3 = 1+8+27 = 36$

Convergence and Divergence

18. A sequence $\{a_n\}$ is said to converge to L if:

1. $a_n$ approaches L as n increases

2. $a_n = L$ for all n

3. $a_n$ is always close to L

4. $|a_n - L| < \epsilon$ for all $n > N$, for any $\epsilon > 0$

Show me the answer

Answer: 4. $|a_n - L| < \epsilon$ for all $n > N$, for any $\epsilon > 0$

Explanation: Formal definition of convergence: A sequence $\{a_n\}$ converges to L if for every $\epsilon > 0$, there exists a natural number N such that:

$|a_n - L| < \epsilon \quad \text{for all } n > N$

Intuitively: $a_n$ gets arbitrarily close to L as n increases.

Notation: $\lim_{n \to \infty} a_n = L$

Example: $a_n = \frac{1}{n}$ converges to 0.

If a sequence does not converge, it diverges.

19. The sequence $a_n = \frac{n^2 + 1}{2n^2 - 3}$ converges to:

1. $0$

2. $\frac{1}{2}$

3. $1$

4. $\infty$

Show me the answer

Answer: 2. $\frac{1}{2}$

Explanation: To find the limit:

Divide numerator and denominator by the highest power of n (which is $n^2$):

As $n \to \infty$: $\frac{1}{n^2} \to 0$ and $\frac{3}{n^2} \to 0$

Therefore:

Alternative method: For rational functions (ratio of polynomials), the limit as $n \to \infty$ is the ratio of leading coefficients: $\frac{1}{2}$.

Tests for Convergence of Series

20. The p-series $\sum \frac{1}{n^p}$ converges if:

1. $p > 0$

2. $p > 1$

3. $p < 1$

4. $p = 1$

Show me the answer

Answer: 2. $p > 1$

Explanation: The p-series:

$\sum_{n=1}^{\infty} \frac{1}{n^p}$

Converges if $p > 1$ Diverges if $p \leq 1$

Important special cases:

• $p = 1$: Harmonic series $\sum \frac{1}{n}$ diverges

• $p = 2$: $\sum \frac{1}{n^2}$ converges (to $\frac{\pi^2}{6}$)

• $p = \frac{1}{2}$: $\sum \frac{1}{\sqrt{n}}$ diverges

The p-series test is a special case of the integral test.

21. A series $\sum a_n$ converges absolutely if:

1. $\sum |a_n|$ converges

2. $\sum a_n$ converges

3. $a_n \to 0$

4. $a_n$ are all positive

Show me the answer

Answer: 1. $\sum |a_n|$ converges

Explanation: Absolute convergence: A series $\sum a_n$ converges absolutely if $\sum |a_n|$ converges.

Important properties:

1. Absolute convergence implies convergence (but not vice versa)

2. Absolutely convergent series can be rearranged without changing the sum

3. Conditionally convergent series (converges but not absolutely) can be rearranged to sum to any real number (Riemann rearrangement theorem)

Example: $\sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n}$ converges conditionally (alternating harmonic series), but $\sum_{n=1}^{\infty} \left|\frac{(-1)^{n+1}}{n}\right| = \sum_{n=1}^{\infty} \frac{1}{n}$ diverges.

Ratio Test

22. For the series $\sum a_n$, if $\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right| = L$, then the series converges absolutely if:

1. $L < 1$

2. $L > 1$

3. $L = 1$

4. $L = 0$

Show me the answer

Answer: 1. $L < 1$

Explanation: Ratio test: Let $L = \lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|$

• If $L < 1$: series converges absolutely

• If $L > 1$: series diverges

• If $L = 1$: test is inconclusive (need another test)

Useful for series with factorials or exponentials.

Example: For $\sum \frac{x^n}{n!}$:

Converges for all x.

Power Series

23. A power series centered at $x_0$ has the form:

1. $\sum a_n (x - x_0)^n$

2. $\sum a_n x^n$

3. $\sum \frac{(x - x_0)^n}{n!}$

4. $\sum \frac{a_n}{n!}$

Show me the answer

Answer: 1. $\sum a_n (x - x_0)^n$

Explanation: A power series centered at $x_0$:

$\sum_{n=0}^{\infty} a_n (x - x_0)^n$

When $x_0 = 0$, it becomes a Maclaurin series: $\sum a_n x^n$

The set of x values for which the series converges is called the interval of convergence.

Within its interval of convergence, a power series defines a function that is infinitely differentiable.

Example: $e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!}$ centered at 0, converges for all x.

Taylor and Maclaurin Series

24. The Maclaurin series expansion of $e^x$ is:

1. $\sum \frac{(-1)^n x^n}{n!}$

2. $\sum \frac{x^n}{n!}$

3. $\sum x^n$

4. $\sum \frac{x^{2n}}{(2n)!}$

Show me the answer

Answer: 2. $\sum \frac{x^n}{n!}$

Explanation: A Maclaurin series is a Taylor series centered at 0.

For $f(x) = e^x$, all derivatives are: $f^{(n)}(x) = e^x$ $f^{(n)}(0) = 1$ for all n

Therefore:

Expanded form: $e^x = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \frac{x^4}{4!} + \cdots$

This series converges for all $x \in \mathbb{R}$.

Euler's formula: $e^{ix} = \cos x + i \sin x$ comes from this series.

25. The radius of convergence R for a power series $\sum a_n (x - x_0)^n$ satisfies:

1. $R = \frac{1}{\limsup |a_n|^{1/n}}$

2. $R = \lim \left|\frac{a_n}{a_{n+1}}\right|$

3. Both 1 and 2 can be used

4. R is always 1

Show me the answer

Answer: 3. Both 1 and 2 can be used

Explanation: The radius of convergence R can be found using:

1. Cauchy-Hadamard theorem: $R = \frac{1}{\limsup_{n \to \infty} |a_n|^{1/n}}$

2. Ratio method (when the limit exists): $R = \lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right|$

Within $|x - x_0| < R$, the series converges absolutely. At $|x - x_0| = R$, convergence needs to be checked separately.

Example: For $\sum \frac{x^n}{n}$:

Using ratio method: $\lim_{n \to \infty} \left|\frac{a_n}{a_{n+1}}\right| = \lim_{n \to \infty} \frac{n+1}{n} = 1$ So $R = 1$.

At $x = 1$: $\sum \frac{1}{n}$ diverges. At $x = -1$: $\sum \frac{(-1)^n}{n}$ converges conditionally.