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Quizzes (518) Questions (9324)

Questions Related to maths

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

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

  1. $\dfrac{-20}{6}$
  2. $\dfrac{-10}{6}$
  3. $\dfrac{-15}{5}$
  4. $\dfrac{-30}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
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 $= \dfrac{-20}{6}$

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

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

  1. $251$
  2. $125$
  3. $215$
  4. $512$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

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

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

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

  1. $210\times { 2 }^{ 17 }$
  2. $420\times { 2 }^{ 17 }$
  3. $420\times { 2 }^{ 87 }$
  4. $210\times { 2 }^{ 87 }$
Reveal answer Fill a bubble to check yourself
C Correct answer
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

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

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?

  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. $q\in \left(\dfrac{22-p}{7}, \dfrac{4(22-p)}{21}\right)$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

This is a mathematical problem involving averages of a sequence. Given the complexity and potential typos in the variable names (n vs a), this is likely a standard inequality problem related to the AM-HM inequality.

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

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

  1. $2^{10}-1$
  2. $2^{10}$
  3. $3^{10}-1$
  4. $3^{10}$
Reveal answer Fill a bubble to check yourself
D Correct answer
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 $3^{10}$