Tag: telescopic summation for infinte series

Questions Related to telescopic summation for infinte series

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
$2\left(\frac{(n-1)!}{n!}\right)= \frac{2}{n}$