Series Summation Methods - Class XI

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

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

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

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

Now,

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.

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,
$a _n = a + (n - 1) d$
$a _{100} = - 5 + (100 - 1)1$
$= - 5 + 99$
$a _{100} =94$
$S _n = \dfrac{100}{2} [-5 + 94]$
$= 50 [89]$
$S _{100} = 4,450$

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

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; $a = 1, d= 2 , a _n = 154$
$153 = 1 + (n - 1) 2$
$153 - 1 = 2n - 2$
$153 - 1 + 2 = 2n$
$n = \dfrac{154}{2} = 77$

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

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
$=(\sqrt 2-1)+(\sqrt 3-\sqrt 2)+.....+4-\sqrt {15}$
$=4-1=3$.

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 } $

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

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]
$= \dfrac{25}{2} [2 + 74]$
$= 12.5 (76)$
$S _n = 950$

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