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Telescopic summation for infinte series - class-XI

Description: telescopic summation for infinte series
Number of Questions: 29
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Tags: binomial theorem, sequence and series maths
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If y= -1  then the value of 
$\displaystyle 1+\frac{1}{y}+\frac{1}{y^{2}}+\frac{1}{y^{3}}+\frac{1}{y^{4}}+\frac{1}{y^{5}}$ is

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. -1

  2. 0

  3. 1

  4. 2

Show answer
Correct Option: B
Explanation:

$1+\frac{1}{y}+\frac{1}{y^{2}}+\frac{1}{y^{3}}+\frac{1}{y^{4}}+\frac{1}{y^{5}}$

Put the value y=-1
Then $1+\frac{1}{-1}+\frac{1}{(-1)^{2}}+\frac{1}{(-1)^{3}}+\frac{1}{(-1)^{4}}+\frac{1}{(-1)^{5}}= 1-1+1-1+1-1= 3-3=0$

Find the sum of the series: 2 + 4 + ......... + 80 using Gauss method, where n = 100

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 4,200

  2. 4,100

  3. 4,300

  4. 4,400

Show answer
Correct Option: B
Explanation:

Sum of 'n' series using Gauss method is
$S _n = \dfrac{n}{2} $  (First term + Last term)
$= \dfrac{100}{2} (2 + 80)$
$= 50 (82)$
$S _{100} = 4,100$

999, 730, 511, 344, 215, ....

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 123

  2. 126

  3. 125

  4. 130

Show answer
Correct Option: B
Explanation:

The rule is $\displaystyle 10^{3}-1$, $\displaystyle 9^{3}+1$, $\displaystyle 8^{3}-1$, $\displaystyle 7^{3}+1$, $\displaystyle 6^{3}-1$, $\displaystyle 5^{3}+1$ The number is 126

Find the Odd one among : 1, 4, 27, 16, 25, 36

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 4

  2. 27

  3. 16

  4. 25

Show answer
Correct Option: B
Explanation:

The numbers are written as $\displaystyle 1^{2}$, $\displaystyle 2^{2}$, $\displaystyle 3^{2}$, $\displaystyle 4^{2}$, $\displaystyle 5^{2}$, $\displaystyle 6^{2}$ etc It should be 9 in place of 27

Find the sum of the series: 1 + 2 + 3 + .......... + 50 using Gauss method.

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 1,275

  2. 1,200

  3. 1,100

  4. 2,000

Show answer
Correct Option: A
Explanation:

Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n(n + 1)}{2}$
$\therefore$ Given n = 50
So, $\dfrac{50(50 + 1)}{2}$
$S _{50} = 1,275$

Find the sum of the following geometric series:
$ \sqrt{7}, \sqrt{21}, 3\sqrt{7},...$ to n terms

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $ \sqrt{7}\left ( \dfrac{3^{-n/2}-1}{\sqrt{3}-1} \right )$

  2. $ \sqrt{6}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

  3. $ \sqrt{7}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

  4. $ \sqrt{5}\left ( \dfrac{3^{n/2}-1}{\sqrt{3}-1} \right )$

Show answer
Correct Option: C
Explanation:

Let $ S _n$ denote the sum of n terms of the G.P. $\sqrt{7}, \sqrt{21}, 3\sqrt{7}, ...,$ 


Cleraly, the given series is a $G.P.$ with first term$=a=\sqrt7$ and common ratio$=r=\sqrt 3$

Then,
$ S _n = \sqrt{7} \left { \dfrac{\left ( \sqrt{3} \right )^{n}-1}{\sqrt{3}-1} \right } = \sqrt{7} \left ( \dfrac{3^{n/2}-1}{3^{1/2}-1} \right )$

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$ is equal to

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $22$

  2. $23$

  3. $44$

  4. $45$

Show answer
Correct Option: B
Explanation:

Now,

${\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;.... + \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\sin^2 46^o+... +\sin^2 98^o)+ \;{\sin ^2}{\text{9}}{{\text{0}}^{\text{o}}}$
$=({\sin ^2}{{\text{2}}^{\text{o}}} + {\sin ^2}{{\text{4}}^{\text{o}}} + \;{\sin ^2}{{\text{6}}^{\text{o}}} + \;...+\sin^2 44^o)+(\cos^2 44^o+... +\cos^2 2^o)+ 1$ [ Since $\sin^2 x^o=\cos^2 (90^o-x^o)$]
$=(\sin^2 2^o+\cos^2 2^o)+(\sin^2 4^o+\cos^2 4^o)+......+(\sin^2 44^o+\cos^2 44^o)+1$
$=1+1+.....+1(22\text{th})+1$
$=22+1$
$=23$.

Find the missing terms :- .........., .........., 8, .........., -1

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 16, 10 and 2

  2. 17, 12 and 3

  3. 17, 12.5 and 3.5

  4. 1, 2 and 3

Show answer
Correct Option: C
Explanation:

Third term, $a _3 = 8$
Fifth term, $a _5 = - 1$
we know that, $a _n = a + (n - 1) d$
$a _3 = a + (3 - 1) d$
$a _3 = a + 2d $     ...... (1)
$a _5 = a + (5 - 1) d$
$-1 = a + 4d $   ....... (2)
comparing equation (1) & (2)
$8    =   a    +   2d$     .... (1)
$-1   =   a    +   4d$     .... (2)
 (+)    (-)    (-)
-------------------
   $a = - 2d$
$d = - 4.5 $
Put $d = -4.5 $ in equation (1)
$8 = a + 2 (- 4.5)$
$8 = a - a$
$a = 17$
$\therefore a _2 = 17 + (2 - 1) d$
$= 17 + 1 (-4.5)$
$a _2 = 12.5$
$a _4 = 17 + (4 - 1) (-4.5)$
$a _4 = 3.5$
The missing terms are 17, 12.5 and 3.5

If the sum of the first n integers is 15. What is n? (use Gauss method)

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 7, 5

  2. 6, -5

  3. 3, -5

  4. 2, 0

Show answer
Correct Option: B
Explanation:

Using Gauss formula,
$S _n = \dfrac{n}{2} (n + 1)$
$15 = \dfrac{n}{2} (n + 1)$
$30 = n (n + 1)$
$30 = n^2 + n$   ....... (1)
By factorization we can write equation (1) as
$n^2 - 6n + 5n - 30 = 0$
$n(n - 6) + 5 (n - 6) = 0$
$n = 6, - 5$
$\therefore$ The n integers will start from -5 to 6.

Find the sum of the first 100 terms -5, -4, -3, -2, -1, 0, 1, 2 ............. using Gauss method

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 4,400

  2. 4,100

  3. 4,200

  4. 4,450

Show answer
Correct Option: D
Explanation:

Given that $a = -5, n = 100, a _n =?, d = 1$
we know that Gauss formula is, $S _n = \dfrac{n}{2}$   [First term + Last term]
To find nth term,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$

Apply Gauss method to find which term of the A.P. 2, 4, 6, 8 ..... is 108?

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 51

  2. 52

  3. 53

  4. 54

Show answer
Correct Option: D
Explanation:

Gauss formula for nth term of the A.P. is $a _n = a + (n - 1) d$
Given: $a = 2, d= 2, a _n = 108$
$108 = 2 + (n - 1) 2$
$108 - 2 = 2n - 2$
$106 + 2 = 2n$
$N = \dfrac{108}{2} = 54$

Find the sum of first 31 terms of an A.P. whose third term is 12 and fourth term is 16.

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 1,983

  2. 1,984

  3. 1,985

  4. 1,986

Show answer
Correct Option: B
Explanation:

Given that, $a _3 = 12; a _4 = 16$
Common difference, $d = a _4 - a _3 = 16 - 12 = 4$
$a _3 - a _2 = d$
$12 - 4 = a _2 $ $\Rightarrow  8$
$d = a _2 - a _1$ 
$a = 4$
We know the formula,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$

The sum of the first 12 terms is 100. The first term is 20. Find the last term. (use Gauss method)

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $\dfrac{-20}{6}$

  2. $\dfrac{-10}{6}$

  3. $\dfrac{-15}{5}$

  4. $\dfrac{-30}{2}$

Show answer
Correct Option: A
Explanation:

Given that $S _n = 100, n = 12, a = 20$. fast term = ?
we know that, $S _n = \dfrac{n}{2}$  [First term + Last term]
$100 = \dfrac{12}{2}$   [20 + Last term]
$100 = 120 + 6 (\text{Last term})$
Last term $= \dfrac{-20}{6}$

Select the correct alternative from the given ones that will complete the series.
$0, 7, 26, 63, 124, ?$

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $251$

  2. $125$

  3. $215$

  4. $512$

Show answer
Correct Option: C
Explanation:

$\underset {(1^{3} - 1)}{0}\rightarrow \underset {(2^{3} - 1)}{7}\rightarrow \underset {(3^{3} - 1)}{26}\rightarrow \underset {(4^{3} - 1)}{63}\rightarrow \underset {(5^{3} - 1)}{124} \rightarrow \underset {(6^{3} - 1)}{215}$.

Find the first term. The sum of the first 100 terms is 1,200. The last term is 150. (use Gauss method)

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 126

  2. -126

  3. 125

  4. -125

Show answer
Correct Option: B
Explanation:

Using Gauss method,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$1,200 = \dfrac{100}{2} [a _1 + 150]$
$1,200 = 50a _1 + 7,500$
$1,200 - 7,500 = 50 a _1$
$a _1 = - 126$

Given $a _1 = 100, a _n = 50$ and $n = 200$. Find their sum using Gauss method.

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 15,000

  2. 14,000

  3. 11,000

  4. 12,000

Show answer
Correct Option: A
Explanation:

Given; $a = 100, a _n = 50, n = 200$
Using Gauss formula,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$= \dfrac{200}{2} [100 + 50]$
$S _n = 15,000$

The famous mathematician associated with finding the sum of the first 100 natural numbers is


telescopic summation for infinte series binomial theorem, sequence and series maths

  1. Pythagoras

  2. Newton

  3. Gauss

  4. Euclid

Show answer
Correct Option: C
Explanation:

Guass found the sum of first 100 natural numbers.

Use Gauss method to find which term of the A.P. 1, 3, 5, 7 ......... is 153?

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 77

  2. 76

  3. 75

  4. 74

Show answer
Correct Option: A
Explanation:

Gauss formula for nth term of the A.P. is $a _n = a + (n - 1) d$
Given; $a = 1, d= 2 , a _n = 154$
$153 = 1 + (n - 1) 2$
$153 - 1 = 2n - 2$
$153 - 1 + 2 = 2n$
$n = \dfrac{154}{2} = 77$

13, 17, 33, 97, 353,....

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 1377

  2. 653

  3. 712

  4. 1273

Show answer
Correct Option: A
Explanation:

Add to each number 4, $\displaystyle 4^{2}$, $\displaystyle 4^{3}$,$\displaystyle 4^{4}$, $\displaystyle 4^{5}$, i.e. $4, 16, 64, 256, 1024$ etc Next number $= 353 + 1024 = 1377$

The value of ${ 1 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 1 }+{ 2 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 2 }+{ 3 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 3 }+.....{ (20) }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 20 }$ is

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $210\times { 2 }^{ 17 }$

  2. $420\times { 2 }^{ 17 }$

  3. $420\times { 2 }^{ 87 }$

  4. $210\times { 2 }^{ 87 }$

Show answer
Correct Option: C
Explanation:

$S={ 1 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 1 }+{ 2 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 2 }+{ 3 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 3 }+.....{ (20) }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 20 }=\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 20 }{ C } } _{ r }$
$=\sum _{ r=1 }^{ 20 }{ { r }^{  } } (r.{ _{  }^{ 20 }{ C } } _{ r })\=20\sum _{ r=1 }^{ 20 }{ { r }^{ 19 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ (r-1+1) } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ { (r-1) }^{  } } .{ _{  }^{ 19 }{ C } } _{ r-1 }+20\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\times 19\sum _{ r=2 }^{ 20 }{ { _{  }^{ 18 }{ C } } _{ r-1 } } +20\times { 2 }^{ 19 }$
$=20\times 19\times { 2 }^{ 18 }+20\times { 2 }^{ 19 }=20\times { 2 }^{ 18 }(19+2)=20\times 21\times { 2 }^{ 18 }=420\times { 2 }^{ 18 }\quad $
(3) option is correct

Let $a-i=i+\dfrac{1}{i}$ for $i=1, 2,..., 20$. Put $p=\dfrac{1}{20}(a _1+n _2+...+n _{20})$ and $q=\dfrac{1}{20}\left(\dfrac{1}{a _1}+\dfrac{1}{a _2}+...+\dfrac{1}{a _{20}}\right)$. Then?

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $q\in \left(0, \dfrac{22-p}{21}\right)$

  2. $q\in \left(\dfrac{22-p}{21}, \dfrac{2(22-p)}{21}\right)$

  3. $q\in \left(\dfrac{2(22-p)}{21}, \dfrac{22-p}{7}\right)$

  4. $q\in \left(\dfrac{22-p}{7}, \dfrac{4(22-p)}{21}\right)$

Show answer
Correct Option: A

The sum $\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $ is equal to 

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $2^{10}-1$

  2. $2^{10}$

  3. $3^{10}-1$

  4. $3^{10}$

Show answer
Correct Option: D
Explanation:
$\displaystyle\sum _{ 0\le i }^{  }{ \sum _{ j\le 10 }^{  }{ \left( _{  }^{ 10 }{ { C } _{ j } } \right) \left( _{  }^{ j }{ { C } _{ i } } \right)  }  } $

$= ^{10}C _{0}(^0C _0) + ^{10}C _1(^1C _1+^1C _0) + ……. + ^{10}C _{10}(^{10}C _{10}+^{10}C _{9}+...…+^{10}C _{0})$

$=2^0.^{10}C _{0} + 2^1.^{10}C _{1} + …….. + 2^{10}.^{10}C _{10}$

Now,


$(1+2x)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2x)^1 + ………….+^{10}C _{10}(2x)^{10} $

Put x = 1,

$(3)^{10} = ^{10}C _{0}(1)^{10} + ^{10}C _1(1)^9(2)^1 + ………….+^{10}C _{10}(2)^{10} $

Thus the required sum is $3^{10}$

What is the value of $\frac {1}{1+\sqrt 2}+\frac {1}{\sqrt 2+\sqrt 3}+\frac {1}{\sqrt 3+\sqrt 4}.....$ upto 15 terms?

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $1$

  2. $2$

  3. $3$

  4. $4$

Show answer
Correct Option: C
Explanation:

By rationalising each term,
$\frac {1}{1+\sqrt 2}=\frac {1}{\sqrt 2+1}\times \frac {\sqrt 2-1}{\sqrt 2-1}=\frac {\sqrt 2-1}{2-1}=\sqrt 2-1$
$\frac

{1}{\sqrt 2+\sqrt 3}=\frac {1}{\sqrt 3+\sqrt 2}\times \frac {\sqrt

3-\sqrt 2}{\sqrt 3-\sqrt 2}=\frac {\sqrt 3-\sqrt 2}{3-2}$
$=\sqrt 3-\sqrt 2$
............................
$\frac {1}{\sqrt {15}+4}=\frac {1}{4+\sqrt {15}}\times \frac {4-\sqrt {15}}{4-\sqrt {15}}=\frac {4-\sqrt {15}}{16-15}$
$=4-\sqrt {15}$
$\therefore$ Given expression
$=(\sqrt 2-1)+(\sqrt 3-\sqrt 2)+.....+4-\sqrt {15}$
$=4-1=3$.

$3, 7, 13, 21, 31, .....$

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $40$

  2. $41$

  3. $42$

  4. $43$

Show answer
Correct Option: D
Explanation:

Add the numbers $4, 6, 8, 10, 12$ etc Next number is $31 +12 = 43$

The sum of infinity of the series $\displaystyle 1+\frac{4}{5}+\frac{7}{5^{2}}+\frac{10}{5^{3}}+$..... is

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $\displaystyle\frac{16}{35}$

  2. $\displaystyle\frac{11}{8}$

  3. $\displaystyle\frac{35}{16}$

  4. $\displaystyle\frac{8}{6}$

Show answer
Correct Option: C
Explanation:

Let $\displaystyle S=1+\frac { 4 }{ 5 } +\frac { 7 }{ { 5 }^{ 2 } } +\frac { 10 }{ { 5 }^{ 3 } } +...$   ...(1)
Multiply (1) by $\displaystyle \frac { 1 }{ 5 } $ , we get

$\displaystyle \frac { 1 }{ 5 } S=\frac { 1 }{ 5 } +\frac { 4 }{ { 5 }^{ 2 } } +\frac { 7 }{ { 5 }^{ 3 } } +\frac { 10 }{ { 5 }^{ 4 } } +...$   ...(2)

$(1) -(2)$, gives 
$\displaystyle \left( 1-\frac { 1 }{ 5 }  \right) S=1+\frac { 3 }{ 5 } +\frac { 3 }{ { 5 }^{ 2 } } +\frac { 3 }{ { 5 }^{ 3 } } +...$

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 5 } \left( 1+\frac { 1 }{ 5 } +\frac { 1 }{ { 5 }^{ 2 } } +... \right) $

$\displaystyle \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 1 }{ 1-\dfrac { 1 }{ 5 }  }  \right) \Rightarrow \dfrac { 4 }{ 5 } S=1+\dfrac { 3 }{ 5 } \left( \dfrac { 5 }{ 4 }  \right) $

$\displaystyle \Rightarrow \frac { 4 }{ 5 } S=1+\frac { 3 }{ 4 } \Rightarrow S=\frac { 35 }{ 16 } $

$1^2+2^2+3^2r^2+4^2r^3+.....$ to $\infty$ is equal to

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $\dfrac{1+r}{(1-r)^2}A$

  2. $\dfrac{1+r}{(1-r)^3}A$

  3. ${1}{(1-r)^3A}$

  4. $\dfrac{1+r}{(1-r)^2}$

Show answer
Correct Option: A
Explanation:

$1+4 + 9r^2 + 16r^3 ....\infty$

$A _2(4 + 16r^3) + (1+9r^2)$
$1 + r + (1 -91)^{-2} = A$
$\dfrac{1+r}{(1-r)^2}A$

Find the sum of the first 25 terms of the A.P.: 2 + 5 + 8 + 11 + ............ (use Gauss method)

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. 910

  2. 930

  3. 950

  4. 940

Show answer
Correct Option: C
Explanation:

Given that $a = 2; d = 3$
$n = 25 ; a _{25} = $?
Using Gauss method, we find the value of $a _{25}$
$a _n = a + (n - 1) d$
$a _{25} = 2 + (25 - 1) 3$
$= 2 + (24) 3$
$a _{25} = 74$
Sum of 'n' series using Gauss method is,
$S _n = \dfrac{n}{2} $   [First term + Last term]
$= \dfrac{25}{2} [2 + 74]$
$= 12.5 (76)$
$S _n = 950$

$73, 71, 67, 61, 59, ....$

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $57$

  2. $55$

  3. $53$

  4. $51$

Show answer
Correct Option: C
Explanation:

Prime numbers in decreasing number series Next number is 53

The value of $ \displaystyle  \left ( 1-\dfrac{1}{3} \right )\left ( 1-\dfrac{1}{4} \right )\left ( 1-\dfrac{1}{5} \right )...\left ( 1-\dfrac{1}{n} \right )  $  is equal to

telescopic summation for infinte series binomial theorem, sequence and series maths

  1. $ \displaystyle \dfrac{1}{n} $

  2. $ \displaystyle \dfrac{2}{n} $

  3. $ \displaystyle \dfrac{3}{n} $

  4. $ \displaystyle \dfrac{4}{n} $

Show answer
Correct Option: B
Explanation:

On simplification of first two term , we observed that
$\left(\frac{2}{3}\right)$ $\times \left(\frac{3}{4}\right)...$
So numerator is the multiplication from 2 to n-1 and denominator is the multiplication from 3 to n
So making it as factorial of first n and n-1 term we have to multiply 1 to numerator and 1 and 2 to denominator
$2\left(\frac{(n-1)!}{n!}\right)= \frac{2}{n}$

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