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

The coefficient of the third term in the Taylor series of  $(x-1)e^x$ is?

  1. $\dfrac{1}{3}$
  2. $3$
  3. $2$
  4. $\dfrac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

We know that taylor series of $e^x$ is given by

$e^x=1+x^1+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...$

$(x-1)e^x=(x-1)(1+x^1+\dfrac{x^2}{2!}+\dfrac{x^3}{3!}+\dfrac{x^4}{4!}...)$

$=-1+\dfrac{1}{2}x^2+\dfrac{1}{3}x^3+\dfrac{1}{8}x^4+\dfrac{1}{30}x^5+....$

Hence, the cofficient of the third term in a taylor series of $(x-1)e^x $ is $\dfrac{1}{3}$

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

The third term in Maclaurin series of $xe^{-x}$ is?

  1. $\dfrac{x^3}{2}$
  2. $\dfrac{x^2}{2}$
  3. $\dfrac{x^3}{3}$
  4. $\dfrac{x}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=xe^{-x}$

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

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

$f^{(0)}(x)=xe^{-x}, \Rightarrow f^{(0)}(0)=0$

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

$f^{(2)}(x)=(x-2)e^{-x}, \Rightarrow f^{(2)}(0)=-2$

$f^{(3)}(x)=(-x+3)e^{-x}, \Rightarrow f^{(3)}(0)=3$

$f^{(4)}(x)=(x-4)e^{-x}, \Rightarrow f^{(4)}(0)=-4$

$f^{(5)}(x)=(-x+5)e^{-x}, \Rightarrow f^{(5)}(0)=5$

$\therefore f(x) \approx \dfrac{0}{0!}x^0+\dfrac{1}{1!}x^1+\dfrac{-2}{2!}x^2+\dfrac{3}{3!}x^3+\dfrac{-4}{4!}x^4+\dfrac{5}{5!}x^5$

$\Rightarrow f(x)\approx x-x^2+\dfrac{1}{2}x^3-\dfrac{1}{6}x^4+\dfrac{1}{24}x^5$

Thus the third term is $\dfrac{1}{2}x^3$.

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

In Maclaurin series of $sin^2x$, the coefficient of the third term is?

  1. $3$
  2. $\dfrac{3}{2}$
  3. $\dfrac{2}{45}$
  4. $\dfrac{2}{65}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

We have to find the coefficient of third term in Maclaurin series of $sin^2 x$.

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=sin^2 x$

Since we have to find the coefficient of the third term, let us take $n=8$.

$\therefore f(x)\approx\sum _{k=0}^{8}\dfrac{f^{(k)}(0)}{k!}x^k$

$f^{(0)}(x)=sin^2 x, \Rightarrow f^{(0)}(0)=0$

$f^{(1)}(x)=2:sinx:cosx, \Rightarrow f^{(1)}(0)=0$

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

$f^{(3)}(x)=-8cosx:sin x, \Rightarrow f^{(3)}(0)=0$

$f^{(4)}(x)=8sin^2x-8cos^2x, \Rightarrow f^{(4)}(0)=-8$

$f^{(5)}(x)=32:sinx:cos x, \Rightarrow f^{(5)}(0)=0$

$f^{(6)}(x)=-32sin^2x+32cos^2x, \Rightarrow f^{(6)}(0)=32$

$f^{(7)}(x)=-128:sinx:cos x, \Rightarrow f^{(7)}(0)=0$

$f^{(8)}(x)=128sin^2x-128cos^2x, \Rightarrow f^{(8)}(0)=-128$

$\therefore f(x) \approx 0x^0+0x^1+\dfrac{2}{2!}x^2+\dfrac{0}{3!}x^3+\dfrac{-8}{4!}x^4+\dfrac{0}{5!}x^5+\dfrac{32}{6!}x^6+\dfrac{0}{7!}x^7+\dfrac{-128}{8!}x^8$

$\Rightarrow f(x)\approx x^2-\dfrac{1}{3}x^4+\dfrac{2}{45}x^6-\dfrac{1}{135}x^5$

Thus the coefficient of third term is $\dfrac{2}{45}$.

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

For Maclaurin series of $log(1+x)$, the coefficient of the third term is given by:

  1. $\dfrac{1}{3}$
  2. $-\dfrac{1}{3}$
  3. $\dfrac{2}{3}$
  4. $\dfrac{-2}{3}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Maclaurin series is given by $f(x)=\sum _{k=0}^{\infty}\dfrac{f^{(k)}(a)}{k!}x^k$ where $a=0$

We have $f(x)=log (1+x)$

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

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

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

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

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

$f^{(3)}(x)=\dfrac{2}{(x+1)^3}, \Rightarrow f^{(3)}(0)=2$

$f^{(4)}(x)=-\dfrac{6}{(x+1)^4}, \Rightarrow f^{(4)}(0)=-6$

$f^{(5)}(x)=\dfrac{24}{(x+1)^5}, \Rightarrow f^{(5)}(0)=24$

$\therefore f(x) \approx \dfrac{0}{0!}x^0+\dfrac{1}{1!}x^1+\dfrac{-1}{2!}x^2+\dfrac{2}{3!}x^3+\dfrac{-6}{4!}x^4+\dfrac{24}{5!}x^5$

$\Rightarrow f(x)\approx x-\dfrac{1}{2}x^2+\dfrac{1}{3}x^3-\dfrac{1}{4}x^4+\dfrac{1}{5}x^5$

Thus the coefficient of third term is $\dfrac{1}{3}$.

Multiple choice physics kinetic theory of gases thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

The SI unit for the coefficient of cubical expansion is

  1. $^\circ C$
  2. $per^\circ C$
  3. $cm^{2}/^\circ C$

  4. none of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
The SI unit of coefficient of cubical expansion is $K^{-1}$
$\gamma =\cfrac { 1 }{ V } \cfrac { dV }{ dT } =\cfrac { 1 }{ { metre }^{ 3 } } \cfrac { { metre }^{ 3 } }{ K } ={ K }^{ -1 }$
Multiple choice business economics and quantitative methods linear correlation spearman's coefficient of correlation spearman's rank correlation method correlation coefficients

If $n=10, \sum x=4,\sum y=3, \sum x^2=8,\sum y^2=9$ and $\sum xy=3,$ then the coefficient of $r _{x,y}$ is

  1. $\frac{3}{4}$
  2. $\frac{1}{5}$
  3. $\frac{1}{6}$
  4. $\frac{1}{4}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Correlation coefficient 
${ r } _{ x,y }=\dfrac { n\sum { xy } -\sum { x } \sum { y }  }{ \sqrt { \left[ n\sum { { x }^{ 2 }-{ \left( \sum { x }  \right)  }^{ 2 } }  \right] \left[ n\sum { { y }^{ 2 }-{ \left( \sum { y }  \right)  }^{ 2 } }  \right]  }  } $

$=\displaystyle\frac { 30-12 }{ \sqrt { 64\times 81 }  } $
$\Rightarrow r _{x,y}=\dfrac{1}{4}$

Multiple choice sum to infinite terms of a gp sequence, progression and series maths

If the expansion in powers of x of the function $\dfrac{1}{(1 - ax)(1 - bx)} , (a \neq b)$ is $a _0 + a _1x + a _2x^2 + .... \, then \, a _n$ is

  1. $\dfrac{b^n - a^n}{b - a}$
  2. $\dfrac{a^n - b^n}{b - a}$
  3. $\dfrac{a^{n+1} - b^{n+1}}{b - a}$
  4. $\dfrac{b^{n+1} - a^{n+1}}{b - a}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
We know $\dfrac{1}{1-ax} = \displaystyle \sum _{r=0}^{\infty} (ax)^r$ 
$1+ax+a^2 x^2+............$
$\dfrac {1}{1-bx} = 1+bx+b^2 x^2 +..........$
So $\dfrac{1}{(1-ax)} \times \dfrac{1}{(1-bx)} = \displaystyle \sum _{r=0}^{\infty} (ax)^r \times \displaystyle \sum _{r=0}^{\infty}(bx)^r$
coefficient of $x^n$ is given by $= a^n + a^{n-1} b + .... + b^n$
given term is $g.p$ will ratio of $b/a$
So $^an = \dfrac{a^n \left(1-\left(\dfrac{b}{a}\right)^{n+1}\right)}{1-b/a} = \dfrac{a^n(a^{n+1} - b^{n+1} )/a^{n+1}}{\dfrac{(a-b)}{a}}$
$= \dfrac{a^{n+1}-b^{n+1}}{a-b}$
$= \dfrac{b^{n+1}-a^{n+1}}{b-a}$
Multiple choice maths geometric sequences sum of terms of g.p sum of n terms of an gp summing geometric series

The sum 
$1 + \left( {1 + x} \right) + \left( {1 + x + {x^2}} \right) + \left( {1 + x + {x^2} + {x^3}} \right) +  \ldots n$  terms equals 

  1. $\frac{{1 - {x^n}}}{{1 - x}}$
  2. $\frac{{x\left( {1 - {x^n}} \right)}}{{1 - x}}$
  3. $\frac{{n\left( {1 - x} \right) - x\left( {1 - {x^n}} \right)}}{{{{\left( {1 - x} \right)}^2}}}$

  4. None of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The sum is S = sum from k=0 to n-1 of (1+x+...+x^k). This is sum from k=0 to n-1 of (1-x^(k+1))/(1-x). Expanding this gives (n - (x + x^2 + ... + x^n)) / (1-x). Using the geometric sum formula for the numerator yields the result in option C.

Multiple choice de moivre’s theorem and its applications demoivre's theorem complex numbers maths

If $C _{o},C _{1},C _{2}...C _{n}$ are the Binomial coefficient in the expansion of $\left ( 1+x \right )^{n}$ then which is not correct

  1. $C _{0}-C _{2}+C _{4}-C _{6}+...=2\tfrac{n}{2}\cos \frac{n\pi }{4}$
  2. $C _{1}-C _{3}+C _{5}+...=2\tfrac{n}{2}\sin \frac{n\pi }{4}$
  3. $C _{1}+C _{5}+C _{9}+C _{13}+...=\tfrac{1}{2}\left ( 2^{n-1}+2\tfrac{n}{2}\sin \frac{n\pi }{4} \right )$

  4. None of these
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

By Binomial theorem
${ \left( 1+x \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }x+{ C } _{ 2 }{ x }^{ 2 }+{ C } _{ 3 }{ x }^{ \ 3 }+{ C } _{ 4 }{ x }^{ 4 }+...$      ...(1)

Substitute $x=i$ in equation (1), we get 
$\Rightarrow { \left( 1+i \right)  }^{ n }={ C } _{ 0 }+{ C } _{ 1 }i-{ C } _{ 2 }-{ C } _{ 3 }i+{ C } _{ 4 }+.....$       ...(2)

Substitute $x=-i$ in equation (1), we get
${ \left( 1-i \right)  }^{ n }={ C } _{ 0 }-{ C } _{ 1 }i-{ C } _{ 2 }+{ C } _{ 3 }i+{ C } _{ 4 }+.....$      ...(3)

Adding (2) & (3), we have
$\Rightarrow 2\left( { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+.... \right) ={ \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....=\dfrac { { \left( 1+i \right)  }^{ n }+{ \left( 1-i \right)  }^{ n } }{ 2 } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }+{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } +\cos { \dfrac { n\pi  }{ 4 }  } -i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2 }  \right) $

$\Rightarrow { C } _{ 0 }{ -C } _{ 2 }{ +C } _{ 4 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\cos { \dfrac { n\pi  }{ 4 }  } $

Subtracting (2) & (3), we have
$\Rightarrow 2i\left( { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.... \right) ={ \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n }$

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....=\dfrac { { \left( 1+i \right)  }^{ n }-{ \left( 1-i \right)  }^{ n } }{ 2i } ={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { { \left( \cos { \dfrac { \pi  }{ 4 }  } +i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n }-{ \left( \cos { \dfrac { \pi  }{ 4 }  } -i\sin { \dfrac { \pi  }{ 4 }  }  \right)  }^{ n } }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+....={ 2 }^{ \dfrac { n }{ 2 }  }\left( \dfrac { \cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 } -\cos { \dfrac { n\pi  }{ 4 }  } +i\sin { \dfrac { n\pi  }{ 4 }  }  }  }{ 2i }  \right) $

$\Rightarrow { C } _{ 1 }{ -C } _{ 3 }{ +C } _{ 5 }+.....={ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $        ...(4)

Substitute $x=1$ in equation (1), we get
$\Rightarrow { C } _{ 0 }+{ C } _{ 1 }{ +C } _{ 2 }+{ C } _{ 3 }+{ C } _{ 4 }+.....\quad ={ 2 }^{ n }$          ...(5)

Substitute $x=-1$ in equation (1), we have
$\ \Rightarrow { C } _{ 0 }{ -C } _{ 1 }{ +C } _{ 2 }-{ C } _{ 3 }+{ C } _{ 4 }+.....\quad =0$       ...(6)

Subtracting (5) & (6), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+..... \right) ={ 2 }^{ n }$
$\Rightarrow { C } _{ 1 }+{ C } _{ 3 }+{ C } _{ 5 }+.....\quad ={ 2 }^{ n-1 }$       ...(7)

Adding (4) & (7), we get
$\Rightarrow 2\left( { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+..... \right) ={ 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  } $

$\Rightarrow { C } _{ 1 }+{ C } _{ 5 }+{ C } _{ 9 }+.....=\dfrac { 1 }{ 2 } \left( { 2 }^{ n-1 }+{ 2 }^{ \dfrac { n }{ 2 }  }\sin { \dfrac { n\pi  }{ 4 }  }  \right) $
Hence, option 'D' is correct.

Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

The sum of $1 + \left( {1 + a} \right)x + \left( {1 + a + {a^2}} \right){x^2} + ....\infty ,\,0 < a,\,x < 1$ equals

  1. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - a} \right)}}$
  2. $\dfrac{1}{{\left( {1 - a} \right)\left( {1 - ax} \right)}}$
  3. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 - ax} \right)}}$
  4. $\dfrac{1}{{\left( {1 - x} \right)\left( {1 + a} \right)}}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$1+\left( 1+a \right) x+\left( 1+a+{ a }^{ 2 } \right) { x }^{ 2 }+\dots \infty $

$ General\quad term\quad of\quad series={ T } _{ n }=\frac { \left( 1-{ a }^{ r+1 } \right)  }{ \left( 1-a \right)  } { x }^{ r }$  

$ { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ \dfrac { \left( 1-{ a }^{ r+1 } \right) { x }^{ r } }{ \left( 1-a \right)  }  } $  

$ \left( 1-a \right) { S } _{ n }=\sum _{ r=1 }^{ \infty  }{ { x }^{ r } } -a\sum _{ r=1 }^{ \infty  }{ \left( ax \right) ^{ r } } $ 

$\left( 1-a \right) { S } _{ n }=\dfrac { 1 }{ 1-x } -\dfrac { a }{ 1-ax } $     ...........  $\because G.P.$ series 

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-ax-a+ax }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ \left( 1-a \right) { S } _{ n }=\dfrac { 1-a }{ \left( 1-x \right) \left( 1-ax \right)  } $  

$ { S } _{ n }=\dfrac { 1 }{ \left( 1-x \right) \left( 1-ax \right)  } $
B is correct
Multiple choice maths geometric sequences introduction to geometric progression understanding geometric progressions introduction to geometric progressions

Coefficient of $x^r$ in $1+(1+x)+(1+x)^2+......+ (1+x)^n$ is 

  1. $^{n+3}C _r$
  2. $^{n+1}C _{r+1}$
  3. $^nC _r$
  4. $^{(n+2)}C _r$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Consider given the series, $1+\left( 1+x \right)+{{\left( 1+x \right)}^{2}}+{{\left( 1+x \right)}^{3}}+...............{{\left( 1+x \right)}^{n}}$

Given series is G.P. which have first term $a=1$,comman ratio $r=1+x$ and number of term $=n$

So sum is,

${{s} _{n}}=\dfrac{a\left( {{r}^{n}}-1 \right)}{r-1}=\dfrac{[{{\left( 1+x \right)}^{n}}-1]}{1+x}={{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$

Now rth term of ${{\left( 1+x \right)}^{n-1}}-{{\left( 1+x \right)}^{-n}}$ is $={}^{n-1}{{C} _{r}}{{x}^{r}}-{}^{-n}{{C} _{r}}.{{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right){{x}^{r}}$


Hence, coefficient of ${{x}^{r}}=\left( {}^{n-1}{{C} _{r}}-{}^{-n}{{C} _{r}} \right)$


Hence, this is the answer.

Multiple choice

What is the Fourier series expansion of a function?

  1. An expansion of a function in terms of a series of sine and cosine functions.

  2. An expansion of a function in terms of a series of exponential functions.

  3. An expansion of a function in terms of a series of polynomial functions.

  4. An expansion of a function in terms of a series of rational functions.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The Fourier series expansion of a function is an expansion of the function in terms of a series of sine and cosine functions.

Multiple choice

What is the coefficient of (x^3) in the expansion of ((1+x)^{10})?

  1. 120

  2. 252

  3. 495

  4. 792
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the binomial theorem, we can expand ((1+x)^{10}) as (\sum_{k=0}^{10} \binom{10}{k} x^k). The coefficient of (x^3) corresponds to (k=3), so we have (\binom{10}{3} = \frac{10!}{3!7!} = \frac{10\cdot9\cdot8}{3\cdot2\cdot1} = 120).

Multiple choice

What is the coefficient of (x^5) in the expansion of ((2x-3)^{8})?

  1. 1344

  2. 2016

  3. 2720

  4. 3584
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using the binomial theorem, we can expand ((2x-3)^{8}) as (\sum_{k=0}^{8} \binom{8}{k} (2x)^{8-k}(-3)^k). The coefficient of (x^5) corresponds to (k=3), so we have (\binom{8}{3} (2)^{8-3}(-3)^3 = \frac{8!}{3!5!} \cdot 2^5 \cdot (-3)^3 = 2720).

Multiple choice

What is the coefficient of (x^6) in the expansion of ((x+2)^{12})?

  1. 924

  2. 1716

  3. 2772

  4. 3003
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the binomial theorem, we can expand ((x+2)^{12}) as (\sum_{k=0}^{12} \binom{12}{k} x^{12-k}2^k). The coefficient of (x^6) corresponds to (k=6), so we have (\binom{12}{6} 2^6 = \frac{12!}{6!6!} \cdot 2^6 = \frac{12\cdot11\cdot10\cdot9\cdot8\cdot7}{6\cdot5\cdot4\cdot3\cdot2\cdot1} \cdot 64 = 924).