# simplification of double summation [closed]

I have solved double summation problem,.Kindly check it whether it is correct or not?? $$\sum_{j=1}^3\sum_{i=1}^j (i+j) = 12$$ thanks

## closed as off-topic by GNUSupporter 8964民主女神 地下教會, Yves Daoust, jvdhooft, A. Goodier, NamasteApr 5 '18 at 11:41

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$$\sum_{j=1}^3\sum_{i=1}^j (i+j)=\sum_{i=1}^1 (i+1)+\sum_{i=1}^2 (i+2)+\sum_{i=1}^3 (i+3)\\=((1+1))+((1+2)+(2+2))+((1+3)+(2+3)+(3+3)).$$