Math Summation Notation for Competitive Exams

Math Summation Notation

In mathematical formula it is indeed the addition of many number or variable which represent to give concise expression for sum of the variable as Sigma or Math Summation Notation.

$Math Summation Notation$

$\sum\limits_{i=1}^n i = \frac{n\left ( n+1 \right )}{2}$

$\sum\limits_{i=1}^n x_{i}$ This expressions shows that sum of values of x, which starting at x₁ and it ending with x_n .
$\sum\limits_{i=1}^n x_{i} = x_{1} + x_{2} + x_{3} +$ … $+ x_{n}$

$\sum\limits_{i=1}^{10} x_{i}$ This expressions shows that sum of values of x, which starting at x₁ and it ending with x₁₀ .

$\sum\limits_{i=1}^{10} x_{i} = x_{1} + x_{2} + x_{3} + x_{4} + x_{5} + x_{6} + x_{7} + x_{8} + x_{9} + x_{10}$

$\sum\limits_{}^{} x = x_{1} + x_{2} + x_{3} +$ … $+ x_{n}$

$\sum\limits_{i=1}^n x_{i}^{2} = x_{1}^{2} + x_{2}^{2} + x_{3}^{2} +$ … $+ x_{n}^{2}$

This expressions shows that sum the squared values of x, which starting at x₁ and it ending with x_n.

Series Summation formula:

$\sum\limits_{i=1}^n i^{1} = \frac{n\cdot \left ( n+1 \right )}{2}$

$\sum\limits_{i=1}^n i^{2} = \frac{n\cdot \left ( n+1 \right )\cdot \left ( 2n+1 \right )}{6}$

$\sum\limits_{i=1}^n i^{3} = \frac{n^{2}\cdot \left ( n+1 \right )^{2}}{4}$

$\sum\limits_{i=1}^n i^{4} = \frac{n\cdot \left ( 2n+1 \right )\cdot \left ( n+1 \right )\cdot \left ( 3n^{2}+3n-1 \right )}{30}$

$\sum\limits_{i=1}^n i^{5} = \frac{n^{2}\cdot \left ( 2n^{2}+2n-1 \right )\cdot\left ( n+1 \right )^{2}}{12}$

$\sum\limits_{i=1}^n i^{6} = \frac{n\cdot \left ( 2n+1 \right )\cdot\left ( n+1 \right )\cdot \left ( 3n^{4}+6n^{3}-3n+1 \right )}{42}$

$\sum\limits_{i=1}^n i^{7} = \frac{n^{2}\cdot \left ( 3n^{4}+6n^{3}-n^{2}-4n+2 \right )\cdot \left ( n+1 \right )^{2}}{24}$

$\sum\limits_{i=1}^n i^{8} = \frac{n\cdot \left ( 2n+1 \right )\cdot \left ( n+1 \right )\cdot \left ( 5n^{6}+15n^{5}+5n^{4}-15n^{3}-n^{2}+9n-3 \right )}{90}$

$\sum\limits_{i=1}^n i^{9} = \frac{n^{2}\cdot \left ( n^{2}+n-1 \right )\cdot \left ( 2n^{4}+4n^{3}-n^{2}-3n+3 \right )\cdot \left ( n+1 \right )^{2}}{20}$

$\sum\limits_{i=1}^n i^{10} = \frac{n\cdot \left ( 2n+1 \right )\cdot \left ( n+1 \right )\cdot \left ( n^{2}+n-1 \right )\cdot \left ( 3n^{6}+9n^{5}+2n^{4}-11n^{3}+3n^{2}+10n-5 \right )}{66}$

* n = Number of terms

* i = Number sequence

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Math Summation Notation

Math Summation Notation

Series Summation formula:

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