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This is the second of our many quizzes on Sigma notation. Click on this to access more such competitive quizzes on Algebra.

Sigma notation Quiz 2

mathematics

 What is the value of $\sum\limits_{n=4}^{7}{\left( {{n}^{2}}+1 \right)}$.

  Substitute n=4 to n=7 in the given expression and simplify.

  $ \text{Given, } \sum\limits_{n=4}^{7}{\left( {{n}^{2}}+1 \right)} \\ \Rightarrow \left( {{4}^{2}}+1 \right)+\left( {{5}^{2}}+1 \right)+\left( {{6}^{2}}+1 \right)+\left( {{7}^{2}}+1 \right) \\ \Rightarrow \left( 16+1 \right)+\left( 25+1 \right)+\left( 36+1 \right)+\left( 49+1 \right) \\ \Rightarrow 130 $

 Find the value of  $2\sum\limits_{n=0}^{8}{\left( {{\left( -1 \right)}^{n}}+1 \right)}$.

 Substitute $n=0$ to $n=8$ in the given expression and simplify.

  $\text{Given, } 2\sum\limits_{n=0}^{8}{\left( {{\left( -1 \right)}^{n}}+1 \right)} \\ \Rightarrow 2\left\{\left( {{\left( -1 \right)}^{0}}+1 \right)+\left( {{\left( -1 \right)}^{1}}+1 \right)+\left( {{\left( -1 \right)}^{2}}+1 \right)+\left( {{\left( -1 \right)}^{3}}+1 \right)+ \\ \left( {{\left( -1 \right)}^{4}}+1 \right)+\left( {{\left( -1 \right)}^{5}}+1 \right)+\left( {{\left( -1 \right)}^{6}}+1 \right)+\left( {{\left( -1 \right)}^{7}}+1 \right)+\left( {{\left( -1 \right)}^{8}}+1 \right)\right\} \\ \Rightarrow 2\left\{ \left(2 \right)+\left( -1+1 \right)+\left( 1+1 \right)+\left( -1+1 \right)+\left( 1+1 \right)+\left( -1+1 \right)+\left( 1+1 \right)+ \\ \left( -1+1 \right)+\left( 1+1 \right) \right\} \\ \Rightarrow 2\left( 2+0+2+0+2+0+2+0+2 \right) \\ \Rightarrow 2\times 10 \\ \Rightarrow 20$

  What will be the sigma notation starting from $n=2$ for the following sum $11+30+67+128+219$.

 Convert the given expression in terms of summation series, then replace it with a summation symbol and $n$.

 Given, $ 11+30+67+128+219 \\ $ This can also be written as,$ \\ \left( {{2}^{3}}+3 \right)+\left( {{3}^{3}}+3 \right)+\left( {{4}^{3}}+3 \right)+\left( {{5}^{3}}+3 \right)+\left( {{6}^{3}}+3 \right) \\ $ We can see that the cube is increasing by 1 and the additional part is common. $ \\ $ So, we can conclude that the equation will be $ \sum\limits_{n=2}^{6}{{{n}^{3}}+3}.$

  What will be the sum of the last three terms of $\sum\limits_{n=0}^{10}{\left( {{n}^{3}}+5 \right)}$.

  Put the value of $n=8,9,10$ in the given equation and add them.

 Given, $\sum\limits_{n=0}^{10}{\left( {{n}^{3}}+5 \right)},\\$ Since we have to find the sum of last three terms which will be $n=8,9,10,$ the given term can be written as, $\sum\limits_{8}^{10}{\left( {{n}^{3}}+5 \right)} \\ \Rightarrow \left( {{8}^{3}}+5 \right)+\left( {{9}^{3}}+5 \right)+\left( {{10}^{3}}+5 \right) \\ \Rightarrow 512+5+729+5+1000+5 \\ \Rightarrow 2256.$

 What will be the difference between  $3\sum\limits_{n=0}^{5}{\left( n+5 \right)\times 8}$ and $2\sum\limits_{n=0}^{5}{\left( n+7 \right)\times 7}$.

 Put the value of $n$ in the given equations separately and then subtract.

$ \text{Given, }3\sum\limits_{n=0}^{5}{\left( n+5 \right)\times 8} \\ \Rightarrow 3\left\{ \left( \left( 0+5 \right) 8 \right)+\left( \left( 1+5 \right)\times 8 \right)+\left( \left( 2+5 \right)\times 8 \right)+\left( \left( 3+5 \right)\times 8 \right)+  \left( \left( 4+5 \right)\times 8 \right)+\left( \left( 5+5 \right)\times 8 \right) \right\} \\ \Rightarrow 3\left( 40+48+56+64+72+80 \right) \\ \Rightarrow 1080 .\\ \text{Given, }2\sum\limits_{n=0}^{5}{\left( n+7 \right)\times 7}  \\ \Rightarrow 2\left\{ \left( \left( 0+7 \right) 7 \right)+\left( \left( 1+7 \right)\times 7 \right)+\left( \left( 2+7 \right)\times 7 \right)+\left( \left( 3+7 \right)\times 7 \right)+  \left( \left( 4+7 \right)\times 7 \right)+\left( \left( 5+7 \right)\times 7 \right) \right\} \\ \Rightarrow 2\left( 49+56+63+70+77+84 \right) \\ \Rightarrow 798 \\  \text{The difference between } 3\sum\limits_{n=0}^{5}{\left( n+5 \right)\times 8} \text{ and }  \\  2\sum\limits_{n=0}^{5}{\left( n+7 \right)\times 7} \text{ is} \\ \Rightarrow 1080-798 = 282.$