Tag: telescopic summation for infinte series

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

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,
$s _n = \dfrac{n}{2} [2a + (n - 1)d]$
$S _{31} = \dfrac{31}{2} [2 \times 4 + (31 - 1)4]$
$= 15.5 [8 + 30 \times 4]$
$=15.5 [128]$
$S _{31} = 1,984$

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 $= \dfrac{-20}{6}$

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

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

Guass found the sum of first 100 natural numbers.

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

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 $= 353 + 1024 = 1377$

$S={ 1 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 1 }+{ 2 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 2 }+{ 3 }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 3 }+.....{ (20) }^{ 2 }.{ _{  }^{ 20 }{ C } } _{ 20 }=\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 20 }{ C } } _{ r }$
$=\sum _{ r=1 }^{ 20 }{ { r }^{  } } (r.{ _{  }^{ 20 }{ C } } _{ r })\=20\sum _{ r=1 }^{ 20 }{ { r }^{ 19 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ (r-1+1) } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\sum _{ r=1 }^{ 20 }{ { (r-1) }^{  } } .{ _{  }^{ 19 }{ C } } _{ r-1 }+20\sum _{ r=1 }^{ 20 }{ { r }^{ 2 } } .{ _{  }^{ 19 }{ C } } _{ r-1 }\=20\times 19\sum _{ r=2 }^{ 20 }{ { _{  }^{ 18 }{ C } } _{ r-1 } } +20\times { 2 }^{ 19 }$
$=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