Tag: telescopic summation for infinte series

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

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

  1. 1,983

  2. 1,984

  3. 1,985

  4. 1,986
Reveal answer Fill a bubble to check yourself
B Correct answer
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,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$

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