Mathematics

Series Expansions and Coefficients

33 Questions

Series expansions involve representing functions as infinite sums, commonly using Taylor and Maclaurin series. This collection focuses on extracting coefficients and expanding binomials. These concepts are frequently tested in mathematics sections of various competitive exams.

Pascal's Triangle coefficientsMaclaurin seriesTaylor series expansionBinomial coefficients

Series Expansions and Coefficients Questions

Multiple choice general knowledge math & puzzles

If the coefficients of (2r + 4)th and (r-2)th terms in the expansion of (1 + x)^18 are equal then the value of r is

  1. 6

  2. 8

  3. 3

  4. 5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

In binomial expansion (1+x)ⁿ, rth term coefficient = nCr-1. For (1+x)¹⁸: (2r+4)th coefficient = 18C(2r+3), (r-2)th coefficient = 18C(r-3). Setting equal: 18C(2r+3) = 18C(r-3). Using nCr = nC(n-r), we get 2r+3 = 18-(r-3) = 21-r, so 3r = 18 and r = 6.

Multiple choice general knowledge math & puzzles

Series expansion of exp(x) is

  1. x-(x^3)/3!+(x^5/5!)-....

  2. 1+x+(x^2)/2!+(x^3)/3!+.....

  3. 1+(x^2)/2!+(x^4)/4!+...

  4. 1+x-(x^2)+(x^3)+....
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The exponential function exp(x) or e^x is represented by the power series 1 + x + x^2/2! + x^3/3! + .... This series includes all powers of x with their respective factorials. Other options represent sin(x), cos(x), or geometric series.

Multiple choice general knowledge math & puzzles

Expansion of cos(x) is

  1. x-(x^3)/3!+(x^5)/5!-....

  2. 1-(x^2/2!)+(x^4/4!)-....

  3. 1+x+(x^2)/2!+(x^3)/3!+...

  4. 1-x+x^2-x^3+....
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Maclaurin series for cos(x) contains only even powers of x with alternating signs, starting with 1. Option A is the series for sin(x) (odd powers only), Option C is e^x (all positive terms), and Option D is a geometric series.

Multiple choice general knowledge math & puzzles

Expansion of cos(x) is

  1. x-(x^3)/3!+(x^5)/5!-....

  2. 1-(x^2/2!)+(x^4/4!)-....

  3. 1+x+(x^2)/2!+(x^3)/3!+...

  4. 1-x+x^2-x^3+....
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Taylor series expansion for cos(x) is 1 - (x^2/2!) + (x^4/4!) - .... It contains only even powers of x and alternating signs. Option 103095 is sin(x), 103097 is exp(x), and 103098 is the expansion for 1/(1+x).

Multiple choice general knowledge math & puzzles

Expansion of 1/(1-x) where x<1 is

  1. 1-(x^2)/2!+(x^4)/4!+...

  2. x-(x^3)/3!+(x^5)/5!-...

  3. 1+x+(x^2)/2!+....

  4. 1+x+(x^2)+...
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The binomial expansion of (1-x)^(-1) gives the infinite geometric series 1 + x + x² + x³ + ... for |x| < 1. Option D correctly shows this pattern without factorials in denominators. Options A and B show trigonometric series (cosine and sine respectively) with factorial denominators, while option C incorrectly includes a factorial denominator in the x² term.

Multiple choice general knowledge math & puzzles

Expansion of exp(-x) is ..

  1. 1+x+(x^2)/2!+(x^3)/3!+..

  2. 1-(x^2)/2!+(x^4)/4!-...

  3. x-(x^3)/3!+(x^5)/5!+..

  4. 1-x+(x^2)/2!-(x^3)/3!+...
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The Taylor series expansion for e^x is 1 + x + x^2/2! + x^3/3!... Substituting -x for x gives 1 - x + x^2/2! - x^3/3! + ..., where signs alternate. Option 108687 correctly shows this alternating pattern.

Multiple choice

The Taylor series expansion of 3 sinx + 2 cos x is
  1. 2 + 3x – x2 – $\frac{x^3}{2}$ + ………
  2. 2 – 3x + x2 – $\frac{x^3}{2}$ + ………
  3. 2 + 3x + x2 + $\frac{x^3}{2}$ + ………
  4. 2 + 3x – x2 + $\frac{x^3}{2}$ + ………
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Multiple choice

The Fourier series expansion of a symmetric and even function, ƒ(x) where f(x) = 1 + $(2x/ \pi), - \pi$< x < 0 and = 1– $(2\pi/\pi)$0 < x < $\pi$ will be
  1. $\sum\limits_{n=1}^\infty (4 / \pi^2n^2)(1 + cos n \pi)$
  2. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 - cos n \pi)$
  3. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 - sin n \pi)$
  4. $\sum\limits_{n=1}^\infty (4 / \pi^2 n^2)(1 + sin n \pi)$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For an even function f(x) = 1 + 2x/π for -π < x < 0 and f(x) = 1 - 2x/π for 0 < x < π, the Fourier cosine coefficients are a₀ = 2 and an = (4/π²n²)(1 - cos nπ). The term (1 - cos nπ) gives 2 for odd n and 0 for even n, resulting in only odd harmonics.

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

In the Taylor series expansion of $\exp \left( x \right) + \sin \left( x \right)$ about the point $x = \pi $, the coefficient of ${\left( {x = \pi } \right)^2}$ is

  1. $\exp \left( \pi \right)$
  2. $0.5\exp \left( \pi \right)$
  3. $\exp \left( \pi \right) + 1$
  4. $\exp \left( \pi \right) - 1$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice taylor's and maclaurin's series applications of differential calculus maths

If $\dfrac{1}{(1-2x)(1+3x)}$ is to be expanded as a power series of $x$, then

  1. $|x|<1/2$
  2. $|x|<1/6$
  3. $\dfrac{1}{3}$
  4. $|x|<1/3$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} \frac { 1 }{ { \left( { 1-2x } \right) \left( { 1+3x } \right)  } }  \ \Rightarrow { \left( { 1-2x } \right) ^{ -1 } }{ \left( { 1+3x } \right) ^{ -1 } }=\frac { 1 }{ 2 } \cdot \frac { 1 }{ 3 } { \left( { \frac { 1 }{ 2 } -x } \right) ^{ -1 } }{ \left( { \frac { 1 }{ 3 } +x } \right) ^{ -1 } } \ \Rightarrow { { by } }\, \, { { using } }\, \, { { binomial } }\, \, { { of } }\, \, { { rational } }\, \, { { index } } \ \Rightarrow { \left( { 1+x } \right) ^{ n } } \ \left| x \right| <1 \ Hence, \ \Rightarrow \left| x \right| <\frac { 1 }{ 2 }  \ \Rightarrow \left| x \right| <\frac { 1 }{ 3 }  \ \therefore \left| x \right| <\frac { 1 }{ 6 }  \end{array}$

Multiple choice taylor's and maclaurin's series applications of differential calculus maths

The coefficient of the fourth term in Taylor series of $x^4 + x ^2-2$ centered at $a=1$.

  1. $4$
  2. $1$
  3. $3$
  4. $6$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Taylor series is given by $f(x)=\sum _{k=0}^{n}\dfrac{f^{(k)}(a)}{k!}(x-a)^k$

We have $f(x)=x^4+x^2-2, a=1$

Since we have to find the coefficient of the fourth term, let us take $n=5$.

$\therefore f(x)\approx\sum _{k=0}^{5}\dfrac{f^{(k)}(1)}{k!}(x-1)^k$

$f^{(0)}(x)=x^4+x^2-2, \Rightarrow f^{(0)}(1)=1+1-2=0$

$f^{(1)}(x)=4x^3+2x, \Rightarrow f^{(1)}(1)=4+2=6$

$f^{(2)}(x)=12x^2+2, \Rightarrow f^{(2)}(1)=12+2=14$

$f^{(3)}(x)=24x, \Rightarrow f^{(3)}(1)=24$

$f^{(4)}(x)=24, \Rightarrow f^{(4)}(1)=24$

$f^{(5)}(x)=0, \Rightarrow f^{(5)}(1)=0$

$\therefore f(x) \approx 0+\dfrac{6}{1!}(x-1)+\dfrac{14}{2!}(x-1)^2+\dfrac{24}{3!}(x-1)^3+\dfrac{24}{4!}(x-1)^4+0$

$\Rightarrow f(x)\approx 6(x-1)+7(x-1)^2+4(x-1)^3+(x-1)^4$

Thus the coefficient of fourth term is $1$.