Can b 0 a 0 + b 1 a 1 + b 2 a 2 + b 3 a 3 + . . . + b n a n be represented as ...Is this correct? (Last step → After taking L.C.M.) b 0 a 0 + b 1 a 1 + b 2 a 2 + b 3 a 3 + . . . + b n a n = ∑ k = 0 n ( b k a k ) = ∑ p = 0 n ( b p a p × ∏ q = 0 n a q ) ∏ q = 0 n a q Thanks!And also feel free to add any extra thing to your answer!

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Can                     b        0                    a        0              +                    b        1                    a        1              +                    b        2                    a        2              +                    b        3                    a        3              +  .  .  .  +                    b        n                    a        n             be represented as ...Is this correct? (Last step    → After taking L.C.M.)                              b          0                          a          0                      +                            b          1                          a          1                      +                            b          2                          a          2                      +                            b          3                          a          3                      +    .    .    .    +                            b          n                          a          n                      =          ∑              k        =        0                    n                            (                                      b          k                          a          k                                    )              =                                                      ∑                              p                =                0                                            n                                                                        (                                                                          b                p                                            a                p                                                          ×                          ∏                              q                =                0                                            n                                                    a              q                                                          )                                                            ∏                          q              =              0                                      n                                            a            q                              Thanks!And also feel free to add any extra thing to your answer!

Brendan Bush · Accepted Answer

This is true simply by linearity of summation: if   k is constant with respect to   p then      ∑          p      =      0        n    k      a    p    =  k  ⋅      ∑          p      =      0        n        a    p  Here,   k  =            ∏              q        =        0            n              a      q      , which is constant with respect to   p, so you can factor it out of the sum and cancel it.You mention LCMs, though it doesn&#039;t look like you used them anywhere. In any case, your equation remains true if you replace             ∏              q        =        0            n              a      q       by       l    c    m    {      a    q    ∣  1  ≤  q  ≤  n  }

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