Mathematics

Series Expansions and Coefficients

41 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

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 maths arithmetic sequences forming an arithmetic progression between two quantities a and b sums arithmetic progression

If  $0 < x , y , a , b < 1 ,$  then the sum of the infinite terms of the series
 $\sqrt { x } ( \sqrt { a } + \sqrt { x } ) + \sqrt { x } ( \sqrt { a b } + \sqrt { x y } ) + \sqrt { x } ( b \sqrt { a } + y \sqrt { x } ) + \ldots$  is

  1. $\dfrac { \sqrt { a x } } { 1 + \sqrt { b } } + \dfrac { x } { 1 + \sqrt { y } }$
  2. $\dfrac { \sqrt { x } } { 1 + \sqrt { b } } + \dfrac { \sqrt { x } } { 1 + \sqrt { y } }$
  3. $\dfrac { \sqrt { x } } { 1 - \sqrt { b } } + \dfrac { \sqrt { x } } { 1 - \sqrt { y } }$
  4. $\dfrac { \sqrt { a x } } { 1 - \sqrt { b } } + \dfrac { x } { 1 - \sqrt { y } }$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The given series can be split into two separate infinite geometric series. The first terms involve powers of b and the second involve powers of y, both with common ratios less than 1. Summing each using the infinite geometric series formula a/(1-r) yields the correct expression.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

Let $a^{k}$ where $k=0.1.2....2013$ are the $2014^{th}$ roots of unity. If $Z _{1}$ and $Z _{2}$ be any two complex number such that $|Z _{1}|=|Z _{2}|=\dfrac{1}{\sqrt{2014}}$, then the value of $\displaystyle \sum _{ k=0 }^{ 2013 }{ { \left| { Z } _{ 1 }+{ a }^{ k }{ Z } _{ 2 } \right|  }^{ 2 } } $ is equal to

  1. $4028$
  2. $0$
  3. $2$
  4. $2014$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Expanding the sum |Z_1 + a^k Z_2|^2 yields sum(|Z_1|^2 + |Z_2|^2 + Z_1 bar(Z_2) a^k + bar(Z_1) Z_2 a^(-k)). Since sum(a^k) for k=0 to n-1 (where n=2014) is 0 for roots of unity, the cross terms vanish. We are left with 2014 * (|Z_1|^2 + |Z_2|^2) = 2014 * (1/2014 + 1/2014) = 2.

Multiple choice finding nth roots of a complex number n th root of unity demoivre's theorem complex numbers maths

If $w \neq 1$ is $n^{th}$ root of unity, then value of $ \displaystyle \sum _{k=0}^{n-1} \left| z _{1} w^{k} z _{2} \right| ^{2}$ is

  1. $n( \left| z _{1} z _{2}\right| ^{2})$
  2. $ \left| z _{1}\right| ^{2}+\left| z _{2}\right| ^{2}$
  3. $( \left| z _{1}\right|+\left| z _{2}\right|) ^{2}$
  4. $n ( \left| z _{1}\right|+\left| z _{2}\right|) ^{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The expression |z1 w^k z2|^2 simplifies to |z1|^2 * |w^k|^2 * |z2|^2. Since |w^k| = 1 for any root of unity, this is |z1|^2 * |z2|^2. Summing this n times yields n * |z1 z2|^2.

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

If $x = 1\, + a + {a^2} + ......\infty $, $y = 1\, + b + {b^2}\,\, + ......\infty $ where $\left| a \right| < 1$ and $\left| b \right| < 1$, then $\left( {1 + ab + {a^2}{b^2} + ........\infty } \right) = ?$

  1. $\frac{xy}{x+y}$
  2. $\frac{x+y}{xy}$
  3. $\frac{xy}{x+y+1}$
  4. $\frac{xy}{x+y-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Given x = 1/(1-a) and y = 1/(1-b), we have a = (x-1)/x and b = (y-1)/y. The series 1 + ab + a^2b^2 + ... is a geometric series with sum 1/(1-ab). Substituting a and b: 1 / (1 - ((x-1)/x)((y-1)/y)) = 1 / (1 - (xy - x - y + 1)/(xy)) = xy / (xy - xy + x + y - 1) = xy / (x + y - 1).

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

If $y=x-x^2+x^3-x^4+....\infty$, then value of x will be?

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

The series y = x - x^2 + x^3 - x^4 + ... is a geometric series with first term a = x and common ratio r = -x. The sum is y = x / (1 - (-x)) = x / (1 + x). Solving for x: y(1 + x) = x, so y + xy = x, which means y = x - xy = x(1 - y). Thus, x = y / (1 - y).

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

If $y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $, then $y =$

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

$y=x^{\dfrac {1}{3}}.x^{\dfrac {1}{9}}.x^{\dfrac {1}{27}}......\infty $
 $ =x^{\cfrac{1}{3}+\cfrac{1}{3^2}+\cfrac{1}{3^3}+........\infty }=x^{\cfrac{1/3}{1-1/3}}=x^{1/2}$
Hence, option 'C' is correct.

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

For first $n$ natural numbers we have the following results with usual notations $ \displaystyle \sum _{r=1}^{n}r =\frac{n(n+1)}{2}, \sum _{r=1}^{n}r^{2} =\frac{n(n+1)(2n+1)}{6},\sum _{r=1}^{n}r^{3}=\left ( \sum _{r=1}^{n}r \right )^{2}$ If $\displaystyle a _{1}a _{2}....a _{n} \in A.P $ then sum to $n$ terms of the sequence $\displaystyle \frac{1}{a _{1}a _{2}},\frac{1}{a _{2}a _{3}},...\frac{1}{a _{n-1}a _{n}}$ is equal to $\displaystyle \frac{n-1}{a _{1}a _{n}}$
 and the sum to $ n$ terms of a $G.P$ with first term '$a$' & common ratio '$r$' is given by  $\displaystyle S _{n}= \frac{lr-a}{r-1}$ for $ r \neq 1 $ for $ r =1 $ sum to $n$ terms of same $G.P.$ is $n$ $a$, where the sum to infinite terms of$G.P.$ is the limiting value of
 $\displaystyle \frac{lr-a}{r-1} $ when $\displaystyle n \rightarrow \infty ,\left |  r \right | < l $ where $l$ is the last term of $G.P.$  On the basis of above data answer the following questionsThe sum to infinite terms of the series $\displaystyle \frac{1}{2}+\frac{1}{6}+\frac{1}{18}+.. $ is equal to ?

  1. $\displaystyle \frac{4}{3}$
  2. $\displaystyle \frac{3}{4}$
  3. $\displaystyle \frac{8}{3}$

  4. Does not exit
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Let, ${ S } _{ \infty  }=\dfrac { 1 }{ 2 } +\dfrac { 1 }{ 6 } +d\frac { 1 }{ 18 } +..\infty $

$\Rightarrow { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( 1+\dfrac { 1 }{ 3 } +\dfrac { 1 }{ { 3 }^{ 2 } } +....\infty  \right) $

As we know that, sum of infinite G.P series $=\dfrac { a }{ 1-r } $

Therefore, $ { S } _{ \infty  }=\dfrac { 1 }{ 2 } \left( \dfrac { 1 }{ 1-\left( 1/3 \right)  }  \right) =\dfrac { 3 }{ 4 } $

Ans: B

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

What is the sum of the infinite geometric series where the beginning term is $2$ and the common ratio is $3$?

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

From the given information, we have
first term $=a=2 $, common ratio $r=3$
We know $S = \dfrac{a}{1-r}$
Therefore, $S = \dfrac{2}{1-3}$
$\Rightarrow S = -1$

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

$S = {3^{10}} + {3^9} + \frac{{{3^9}}}{4} + \frac{{{3^7}}}{2} + \frac{{{{5.3}^6}}}{{16}} + \frac{{{3^2}}}{{16}} + \frac{{{{7.3}^4}}}{{64}} + .........$ upto infinite terms, then $\left( {\frac{{25}}{{36}}} \right)S$ equal to 

  1. ${6^9}$
  2. ${3^{10}}$
  3. ${3^{11}}$
  4. ${2.3^{10}}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l}S = {3^{10}} + {3^9} + \cfrac{{{3^9}}}{4} + \cfrac{{{3^7}}}{2} + \cfrac{{{{5.3}^6}}}{{16}} + \cfrac{{{3^2}}}{{16}} + \cfrac{{{{7.3}^4}}}{{64}} + .........\infty \S = \cfrac{{1 \times {3^{10}}}}{{{2^0}}} + \cfrac{{2 \times {3^9}}}{{{2^1}}} + \cfrac{{3 \times {3^8}}}{{{2^2}}} + \cfrac{{4 \times {3^7}}}{{{2^3}}} + \cfrac{{5 \times {3^6}}}{{{2^4}}} + \cfrac{{6 \times {3^5}}}{{{2^5}}} + \cfrac{{7 \times {3^4}}}{{{2^6}}} + .....\infty \\cfrac{S}{6} = \cfrac{{{3^9}}}{2} + \cfrac{{2 \times {3^8}}}{{{2^2}}} + .......\infty \S - \cfrac{S}{6} = \cfrac{{{3^{10}}}}{2^0} + \cfrac{{{3^9}}}{{{2^1}}} + \cfrac{{{3^8}}}{{{2^2}}} + ........\infty \\cfrac{{6S - S}}{6} = \cfrac{{{3^{10}}}}{{1 - \cfrac{1}{6}}} = \cfrac{{{3^{10}}}}{5}\left( 6 \right) \end{array}$

$\dfrac56S=\dfrac{3^{10}\times 6 }{5}$
$\therefore \dfrac {25}{36}S=3^{10}$

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