I'm reading my vector calculus text when I encountered below formula. ( ∑ i = 1 n x i ) 2 = ( ∑ i = 1 n x i 2 + ∑ i &lt; j 2 x i x j )

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I&#039;m reading my vector calculus text when I encountered below formula.  (      ∑          i      =      1        n              x              i                  )    2    =  (      ∑          i      =      1        n              x              i            2        +      ∑          i      &amp;lt;      j            2          x              i                    x              j              )

kpgt1z · Accepted Answer

Consider the array      [                                                      x              1              2                                                                          x              1                                      x              2                                                                          x              1                                      x              3                                                            x            1                                x            4                                                    …                                      x            1                                x            n                                                                          x              1                                      x              2                                                                          x              2              2                                                            x            2                                x            3                                                x            2                                x            4                                                    …                                      x            2                                x            n                                                                          x              1                                      x              3                                                            x            2                                x            3                                                              x              3              2                                                            x            3                                x            4                                                    …                                      x            3                                x            n                                                            x            1                                x            4                                                x            2                                x            4                                                x            3                                x            4                                                              x              4              2                                                            x            4                                x            5                                            …                                      x            4                                x            n                                                                        ⋮                                      ⋱                                                                        ⋮                                                              ⋮                                                  x            1                                x            n                                                x            2                                x            n                                    …                                                  …                                                    x              n              2                                            ]    .This is the array consisting of all terms in the expansion of   (      x    1    +  …  +      x    n        )    2  . Notice that the array is symmetric about the diagonal. So therefore we just need to notice that                                          (                          ∑                              i                =                1                            n                                      x              i                        )                    2                =        (                  x          1                +        …        +                  x          n                          )          2                                    =                    (sum of all terms along the diagonal)                                                +                  (                                    t              w              i              c              e                                 the sum of all terms above the diagonal)                                                  =                              ∑                      i            =            1                    n                          x          i          2                +        2                  ∑                      i            &amp;lt;            j                                    x          i                          x          j                .

I'm reading my vector calculus text when I encountered below formula. ( <munderover> <mo m

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