Maths > Sequence and Series > 6.0 $\Sigma $, Sigma Notation
Sequence and Series
1.0 Introduction
2.0 Arithmetic Sequence or Arithmetic Progression (A.P.)
3.0 Geometric Sequence or Geometric Progression (G.P.)
4.0 Harmonic Sequence or Harmonic Progression (H.P.)
5.0 Relation among A.M., G.M. and H.M.
6.0 $\Sigma $, Sigma Notation
7.0 Arithmetico-Geometric Series (A.G.S.)
8.0 Miscellaneous Series
6.1 Important results:
(a) Sum of first $n$ natural numbers:
$$\begin{equation} \begin{aligned} \sum {n = 1 + 2 + 3 + ... + n} \\ \sum {n = \frac{{n(n + 1)}}{2}} \\\end{aligned} \end{equation} $$
(b) Sum of squares of first $n$ natural numbers:
$$\begin{equation} \begin{aligned} \sum {{n^2} = {1^2} + {2^2} + {3^2} + ... + {n^2}} \\ \sum {{n^2} = \frac{{n(n + 1)(2n + 1)}}{6}} \\\end{aligned} \end{equation} $$
Proof: $$\because {(x + 1)^3} - {x^3} = 3{x^2} + 3x + 1$$
Putting $x = 1,2,3,4,...,n$ then
$$\begin{equation} \begin{aligned} {2^3} - {1^3} = {3.1^2} + 3.1 + 1 \\ {3^3} - {2^3} = {3.2^2} + 3.2 + 1 \\ {4^3} - {3^3} = {3.3^2} + 3.3 + 1 \\ .................................. \\ {(n + 1)^3} - {n^3} = 3{n^2} + 3n + 1 \\\end{aligned} \end{equation} $$
Adding all we get, $$\begin{equation} \begin{aligned} {(n + 1)^3} - {1^3} = 3({1^2} + {2^2} + {3^2} + ... + {n^2}) + 3(1 + 2 + 3 + ... + n) + (1 + 1 + 1... + n{\text{ times)}} \\ \Rightarrow {n^3} + 3{n^2} + 3n = 3\sum {{n^2} + 3\sum n } + n \\ 3\sum {{n^2} = } {\text{ }}{n^3} + 3{n^2} + 3n - 3\frac{{n(n + 1)}}{2} - n \\ \sum {{n^2} = } \frac{{n(n + 1)(2n + 1)}}{6} \\\end{aligned} \end{equation} $$
(c) Sum of cubes of first $n$ natural numbers:
$$\begin{equation} \begin{aligned} \sum {{n^3} = {1^3} + {2^3} + {3^3} + ... + {n^3}} \\ {\sum {{n^3} = \left[ {\frac{{n(n + 1)}}{2}} \right]} ^2} = {\left[ {\sum n } \right]^2} \\\end{aligned} \end{equation} $$
Proof: $$\because {(x + 1)^4} - {x^4} = 4{x^3} + 6{x^2} + 4x + 1$$
Putting $x = 1,2,3,4,...,n$ then
$$\begin{equation} \begin{aligned} {2^4} - {1^4} = {4.1^3} + {6.1^2} + 4.1 + 1 \\ {3^4} - {2^4} = {4.2^3} + {6.2^2} + 4.2 + 1 \\ ............................................. \\ {(n + 1)^4} - {n^4} = 4.{n^3} + 6.{n^2} + 4.n + 1 \\\end{aligned} \end{equation} $$
Adding all, we get
$$\begin{equation} \begin{aligned} {(n + 1)^4} - {1^4} = 4({1^3} + {2^3} + {3^3} + ... + {n^3}) + 6({1^2} + {2^2} + {3^2} + ... + {n^2}) + 4(1 + 2 + 3 + ... + n) + (1 + 1 + 1 + ... + n{\text{ times)}} \\ {n^4} + 4{n^3} + 6{n^2} + 4n = 4\sum {{n^3} + 6\sum {{n^2} + 4\sum n + n} } \\ 4\sum {{n^3} = {n^2}{{(n + 1)}^2}} \\ \sum {{n^3} = } {\left[ {\frac{{n(n + 1)}}{2}} \right]^2} \\\end{aligned} \end{equation} $$
Note: If $n$th term of the series is given by ${T_n} = a{n^3} + b{n^2} + cn + d$ where $a$, $b$, $c$ and $d$ are constants, then sum of $n$ terms i.e., $${S_n} = \sum {{T_n}} = a\sum {{n^3} + b\sum {{n^2} + c\sum n } + \sum d } $$
