Square of a generic sum of elements

The formula for computing the square of a generic sum of terms is as follows:

\\left(\\sum_{k=1}^{K}a_{k}\\right)^{2}=\\sum_{k=1}^{K}a_{k}^{2}+2\\sum_{j=1}^{K}%\\sum_{i<j}a_{i}a_{j}

We can prove this property by induction, considering that it holds for K=2, since

(a+b)^{2}=a^{2}+b^{2}+2ab

and that if the property holds for a generic K, then it holds also for K+1, as is proven in the following passages:

\\aligned\\left(\\sum_{k=1}^{K+1}a_{k}\\right)^{2}&=\\left(\\sum_{k=1}^{K}a_{k}+a_{K%+1}\\right)^{2}=\\left(\\sum_{k=1}^{K+1}a_{k}\\right)^{2}+a_{K+1}^{2}+2\\sum_{k=1}^%{K}a_{k}a_{K+1}\\\\&=\\sum_{k=1}^{K}a_{k}^{2}+2\\sum_{j=1}^{K}\\sum_{i<j}a_{i}a_{j}+a_{K+1}^{2}+2%\\sum_{k=1}^{K}a_{k}a_{K+1}=\\\\&=\\sum_{k=1}^{K+1}a_{k}^{2}+2\\sum_{j=1}^{K+1}\\sum_{i<j}a_{i}a_{j}

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