Sigma Notation
Complete the following sums by filling in the missing sections appropriately.
   
4(1)2+4(2)2+4(3)2+⋯+4(11)2=\displaystyle {4}{\left({1}\right)}^{{2}}+{4}{\left({2}\right)}^{{2}}+{4}{\left({3}\right)}^{{2}}+\cdots+{4}{\left({11}\right)}^{{2}}= ∑\displaystyle \sum  
  n=1\displaystyle {n}={1}  
   
2(2+3)2+3(3+3)2+4(4+3)2+⋯+8(8+3)2=\displaystyle {2}{\left({2}+{3}\right)}^{{2}}+{3}{\left({3}+{3}\right)}^{{2}}+{4}{\left({4}+{3}\right)}^{{2}}+\cdots+{8}{\left({8}+{3}\right)}^{{2}}= ∑\displaystyle \sum  
  n=2\displaystyle {n}={2}  
   
4131+4232+4333+⋯+410310=\displaystyle \frac{{{4}^{{1}}}}{{{3}^{{1}}}}+\frac{{{4}^{{2}}}}{{{3}^{{2}}}}+\frac{{{4}^{{3}}}}{{{3}^{{3}}}}+\cdots+\frac{{{4}^{{10}}}}{{{3}^{{10}}}}= ∑\displaystyle \sum  
  n=1\displaystyle {n}={1}  
   
−1(3)−2+−1(3)−1+−1(3)0+⋯+−1(3)12=\displaystyle -{1}{\left({3}\right)}^{{-{{2}}}}+-{1}{\left({3}\right)}^{{-{{1}}}}+-{1}{\left({3}\right)}^{{0}}+\cdots+-{1}{\left({3}\right)}^{{12}}= ∑\displaystyle \sum  
  n=−2\displaystyle {n}=-{2}  

 

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