# sankalan

sankhyaaon ke kisi kram ko jodne ki sankriya sankalan (Summation) kahalaati hai. iska parinaam yog (sum) ya kulayog (total) kahalaati hai.

## prateek (notation)

### kaipitl sigma (Capital-sigma)

yeh nimnalikhit tareeke se paribhaashit hai-

${\displaystyle \sum _{i=m}^{n}x_{i}=x_{m}+x_{m+1}+x_{m+2}+\cdots +x_{n-1}+x_{n}.}$
ek udaaharan-
${\displaystyle \sum _{k=2}^{6}k^{2}=2^{2}+3^{2}+4^{2}+5^{2}+6^{2}=90.}$

## sankalan se sambandhit sarvasamikaaeain (Identities)

### saamaanya

${\displaystyle \sum _{n=s}^{t}C\cdot f(n)=C\cdot \sum _{n=s}^{t}f(n)}$, where C is a constant
${\displaystyle \sum _{n=s}^{t}f(n)+\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)+g(n)\right]}$
${\displaystyle \sum _{n=s}^{t}f(n)-\sum _{n=s}^{t}g(n)=\sum _{n=s}^{t}\left[f(n)-g(n)\right]}$
${\displaystyle \sum _{n=s}^{t}f(n)=\sum _{n=s+p}^{t+p}f(n-p)}$
${\displaystyle \sum _{n=s}^{j}f(n)+\sum _{n=j+1}^{t}f(n)=\sum _{n=s}^{t}f(n)}$
${\displaystyle \left(\sum _{i=k_{0}}^{k_{1}}a_{i}\right)\left(\sum _{j=l_{0}}^{l_{1}}b_{j}\right)=\sum _{i=k_{0}}^{k_{1}}\sum _{j=l_{0}}^{l_{1}}a_{i}b_{j}}$
${\displaystyle \sum _{i=k_{0}}^{k_{1}}\sum _{j=l_{0}}^{l_{1}}a_{i,j}=\sum _{j=l_{0}}^{l_{1}}\sum _{i=k_{0}}^{k_{1}}a_{i,j}}$
${\displaystyle \sum _{n=0}^{t}f(2n)+\sum _{n=0}^{t}f(2n+1)=\sum _{n=0}^{2t+1}f(n)}$
${\displaystyle \sum _{n=0}^{t}\sum _{i=0}^{z-1}f(z\cdot n+i)=\sum _{n=0}^{z\cdot t+z-1}f(n)}$
${\displaystyle \sum _{n=s}^{t}\ln f(n)=\ln \prod _{n=s}^{t}f(n)}$
${\displaystyle c^{\left[\sum _{n=s}^{t}f(n)\right]}=\prod _{n=s}^{t}c^{f(n)}}$

${\displaystyle \sum _{i=m}^{n}1=n-m+1}$
${\displaystyle \sum _{i=1}^{n}{\frac {1}{i}}=H_{n}}$ (See Harmonic number)
${\displaystyle \sum _{i=m}^{n}i={\frac {(n-m+1)(n+m)}{2}}}$ (see arithmetic series)
${\displaystyle \sum _{i=0}^{n}i=\sum _{i=1}^{n}i={\frac {n(n+1)}{2}}}$ (Special case of the arithmetic series)
${\displaystyle \sum _{i=1}^{n}i^{2}={\frac {n(n+1)(2n+1)}{6}}={\frac {n^{3}}{3}}+{\frac {n^{2}}{2}}+{\frac {n}{6}}}$
${\displaystyle \sum _{i=1}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}={\frac {n^{4}}{4}}+{\frac {n^{3}}{2}}+{\frac {n^{2}}{4}}=\left[\sum _{i=1}^{n}i\right]^{2}}$
${\displaystyle \sum _{i=1}^{n}i^{4}={\frac {n(n+1)(2n+1)(3n^{2}+3n-1)}{30}}={\frac {n^{5}}{5}}+{\frac {n^{4}}{2}}+{\frac {n^{3}}{3}}-{\frac {n}{30}}}$
${\displaystyle \sum _{i=0}^{n}i^{p}={\frac {(n+1)^{p+1}}{p+1}}+\sum _{k=1}^{p}{\frac {B_{k}}{p-k+1}}{p \choose k}(n+1)^{p-k+1}}$ where ${\displaystyle B_{k}}$ denotes a Bernoulli number

The following formulas are manipulations of ${\displaystyle \sum _{i=1}^{n}i^{3}=\left(\sum _{i=1}^{n}i\right)^{2}}$ generalized to begin a series at any natural number value (i.e., ${\displaystyle m\in \mathbb {N} }$):

${\displaystyle \left(\sum _{i=m}^{n}i\right)^{2}=\sum _{i=m}^{n}(i^{3}-im(m-1))}$
${\displaystyle \sum _{i=m}^{n}i^{3}=\left(\sum _{i=m}^{n}i\right)^{2}+m(m-1)\sum _{i=m}^{n}i}$

In the summations below x is a constant not equal to 1

${\displaystyle \sum _{i=m}^{n-1}x^{i}={\frac {x^{m}-x^{n}}{1-x}}}$ (m < n; see geometric series)
${\displaystyle \sum _{i=0}^{n-1}x^{i}={\frac {1-x^{n}}{1-x}}}$ (geometric series starting at 1)
${\displaystyle \sum _{i=0}^{n-1}ix^{i}={\frac {x-nx^{n}+(n-1)x^{n+1}}{(1-x)^{2}}}}$
${\displaystyle \sum _{i=0}^{n-1}i2^{i}=2+(n-2)2^{n}}$ (special case when x = 2)
${\displaystyle \sum _{i=0}^{n-1}{\frac {i}{2^{i}}}=2-{\frac {n+1}{2^{n-1}}}}$ (special case when x = 1/2)

### dvipd gunaankon vaale sankalan (summations involving binomial coefficients)

There exist enormously many summation identities involving binomial coefficients (a whole chapter of Concrete Mathematics is devoted to just the basic techniques). Some of the most basic ones are the following.

${\displaystyle \sum _{i=0}^{n}{n \choose i}=2^{n}}$
${\displaystyle \sum _{i=1}^{n}i{n \choose i}=n2^{n-1}}$
${\displaystyle \sum _{i=0}^{n}i!\cdot {n \choose i}=\lfloor n!\cdot e\rfloor }$
${\displaystyle \sum _{i=0}^{n-1}{i \choose k}={n \choose k+1}}$
${\displaystyle \sum _{i=0}^{n}{n \choose i}a^{(n-i)}b^{i}=(a+b)^{n}}$, the binomial theorem

## vruddhi dar

The following are useful approximations (using theta notation):

${\displaystyle \sum _{i=1}^{n}i^{c}=\Theta (n^{c+1})}$ for real c greater than −1
${\displaystyle \sum _{i=1}^{n}{\frac {1}{i}}=\Theta (\log n)}$ (See Harmonic number)
${\displaystyle \sum _{i=1}^{n}c^{i}=\Theta (c^{n})}$ for real c greater than 1
${\displaystyle \sum _{i=1}^{n}\log(i)^{c}=\Theta (n\cdot \log(n)^{c})}$ for non-negative real c
${\displaystyle \sum _{i=1}^{n}\log(i)^{c}\cdot i^{d}=\Theta (n^{d+1}\cdot \log(n)^{c})}$ for non-negative real c, d
${\displaystyle \sum _{i=1}^{n}\log(i)^{c}\cdot i^{d}\cdot b^{i}=\Theta (n^{d}\cdot \log(n)^{c}\cdot b^{n})}$ for non-negative real b > 1, c, d