1. The series \(1 + 2 + 3 +…n\) is written as \(\sum n \) and it is equal to \(\frac{{n(n + 1)}}{2}\)
2. \(\sum {\left( {\sum n } \right) = \sum {1 + \sum {2 + \sum {3 + ....\sum n } } } } \) 
   \(= (1) + (1 + 2) + (1 + 2 + 3) + ……. n\) terms 
   =\(\left( {\frac{{n(n + 1)(n + 2)}}{{3\,\, \times \,\,2\,\, \times \,\,1}}} \right)\) 
   Note: We can generalize this formula. Let us write the summation 
   \(\sum n  = {S^1}(n),\;\)\(\sum {\left( {\sum n } \right) = {S^2}(n)} \) and
   \(\sum {\left( {\sum {......} r\;times} \right) = {S^r}(n)} \), then
   \({S^r}(n) = {\;^{n + r}}{C_{r + 1}}\). By using this formula we have \({S^4}(n) = \frac{{(n + 4)(n + 3)(n + 2)(n + 1)n}}{{5.4.3.2.1}}\)
3. Sum of the first \(n\) odd numbers is \({n^2}\) and sum of the first \(n\) even numbers is \(n(n + 1)\). 
   \(1 + 3 + 5 + …… n\) terms \(= {n^2}\) 
   \(2 + 4 + 6 + …… n\) terms = \(n(n + 1)\)
4. Sum of the squares of the first n natural numbers is\(\frac{{n(n + 1)(2n + 1)}}{6}\)and sum of the cubes of the first \(n\) natural numbers is \({\left\{ {\frac{{n(n + 1)}}{2}} \right\}^2}\) 
   \(\sum {{n^2} = {1^2} + {2^2} + ....{n^2} = \frac{{n(n + 1)(2n + 1)}}{6}} \) 
   \(\sum {{n^3} = {1^3} + {2^3} + .....n\,\,terms = {{\left( {\frac{{n(n + 1)}}{2}} \right)}^2}} \) 
   \(\sum {{n^4}}  = \left\{ {\frac{{n(n + 1)(2n + 1)(3{n^2} + 3n - 1)}}{{30}}} \right\}\)
5. \(1.2 + 2.3 + 3.4 + \,.....n\,terms\,\,\,\,\,\,\, = \frac{{n(n + 1)(n + 2)}}{3}\) 
   \(1.2.3 + 2.3.4 + 3.4.5.....n\;terms\,\,\,\,\, = \frac{{n(n + 1)(n + 2)(n + 3)}}{4}\)
6. Some times \({n^{th}}\) term of the series is a polynomial in \(n\), suppose \({n^{th}}\) term of a series is:  
   \({T_n} = {\rm{ }}a{n^3} + b{n^2} + cn + d\) 
   The sum of \(n\) terms is given by  
   \({S_n} = \sum {{T_n} = a\sum {{n^3} + b\sum {{n^2}}  + c\sum {n + \sum d } } } \) 
   For example suppose the series is: 
   \(\left( {{1^2} + {\rm{ }}2} \right){\rm{ }} + {\rm{ }}\left( {{2^2} + {\rm{ }}4} \right){\rm{ }} + {\rm{ }}\left( {{3^2} + {\rm{ }}6} \right){\rm{ }} + {\rm{ }} \ldots ..n\) terms 
   \({n^{th}}\) term of the series is \({n^2} + {\rm{ }}2n\), hence sum of the series is given by \(\sum {{n^2}}  + \sum {2n} \) 
   = \(\frac{{n(n + 1)(2n + 1)}}{6} + n(n + 1) = \frac{{n(n + 1)(2n + 7)}}{6}\)

Example 1: What is the \({305^{th}}\) term of the sequence 

\(a,{\rm{ }}b,{\rm{ }}b,{\rm{ }}c,{\rm{ }}c,{\rm{ }}c,{\rm{ }}d,{\rm{ }}d,{\rm{ }}d,{\rm{ }}d, \ldots  \ldots  \ldots z,{\rm{ }}z,{\rm{ }}{z_{(26{\rm{ }}times)}}?\) 

Solution: Suppose up to 300 terms a total of  \(\lambda \) alphabets are used. Since \(a\) is used once, \(b\) is used twice and so on, then \(\frac{{\lambda (\lambda  + 1)}}{2} \ge 305\) \( \Rightarrow \lambda  > {\rm{ }}24\). Thus \({305^{th}}\) term of this sequence will be 25 letter i.e. \(y\). Note that up to 24th alphabet (i.e. \(x\)) a total of 300 terms are formed, for the next 25 terms \(y\) is repeated, thus at 305th term \(y\) appears. 

Example 2: Find the sum of the series

\(\frac{1}{{\sqrt 1  + \sqrt 3 }} + \frac{1}{{\sqrt 3  + \sqrt 5 }} + \frac{1}{{\sqrt 5  + \sqrt 7 }} + .......\frac{1}{{\sqrt {79}  + \sqrt {81} }}\)

Solution: The first term of the series can be written as: \(\frac{{\sqrt 3  - \sqrt 1 }}{2}\), similarly second term of the series can be written as \(\frac{{\sqrt 5  - \sqrt 3 }}{2}\) and so on. Thus the sum of the series is:

\(\frac{1}{2}\left\{ {\left( {\sqrt 3  - \sqrt 1 } \right) + \left( {\sqrt 5  - \sqrt 3 } \right) + .....\sqrt {81}  - \sqrt {79} } \right\} = \frac{1}{2}\left( {9 - 1} \right) = 4\) 

Example 3: Find the sum of the series:

\({1^3} - {2^3} + {3^3} - {4^3} + {5^3} - ...... + {19^3} - {20^3}\)

Solution: The given series can be written as:

\(\left( {{1^3} + {2^3} + {3^3} + {{...20}^3}} \right) - 2\left( {{2^3} + {4^3} + {6^3} + {{..20}^3}} \right)\)

=  \(\left( {{1^3} + {2^3} + {3^3} + {{...20}^3}} \right) - 2 \times {2^3}\left( {{1^3} + {2^3} + {3^3} + {{..10}^3}} \right)\)

= \({\left( {\frac{{20 \times 21}}{2}} \right)^2} - 16 \times {\left( {\frac{{10 \times 11}}{2}} \right)^2}= - 4300\).

Example 4: Find the sum of the series

\(1 \times 50{\rm{ }} + {\rm{ }}2 \times 49{\rm{ }} + {\rm{ }}3 \times 48{\rm{ }} + {\rm{ }} \ldots .50 \times 1\)

Solution: nth term of the series can be written as:

\({T_n} = n \times \left( {51{\rm{ }}--n} \right){\rm{ }} = {\rm{ }}51n--{n^2}\)

Hence sum up to n terms = \(51\sum n  - \sum {{n^2}} \)

= \(51\left( {\frac{{n(n + 1)}}{2}} \right) - \left( {\frac{{n(n + 1)(2n + 1)}}{6}} \right)\)

= \(51\left( {\frac{{50(51)}}{2}} \right) - \left( {\frac{{50(51)(101)}}{6}} \right) = 22100\)