How to find the values of these kind of summations: ∑ i = 0 6 ( 6 − i ) ∗ ∑ j = 1 6 ( 7 − j ) ∗ ∑ k = 2 7 ( 8 − k ) ∗ ∑ ℓ = 3 8 ( 9 − ℓ )

Question

How to find the values of these kind of summations:            ∑              i        =        0            6        (    6    −    i    )        ∗              ∑              j        =        1            6        (    7    −    j    )        ∗              ∑              k        =        2            7        (    8    −    k    )        ∗              ∑              ℓ        =        3            8        (    9    −    ℓ    )

Jaruckigh · Accepted Answer

Use that                              ∑                      t            =            a                    b                (        c        −        t        )                            =                  (                      ∑                          t              =              a                        b                    c          )                −                  (                      ∑                          t              =              a                        b                    t          )                                                                =                  (                      ∑                          t              =              a                        b                    c          )                −                  (                      ∑                          s              =              0                                      b              −              a                                (          s          +          a          )          )                                                                =                  (                      ∑                          t              =              a                        b                    c          )                −                  (                      ∑                          s              =              0                                      b              −              a                                s          )                −                  (                      ∑                          s              =              0                                      b              −              a                                a          )                                                                =        (        b        −        a        +        1        )        c        −        (        b        −        a        +        1        )        a        −                  (                      ∑                          s              =              0                                      b              −              a                                s          )                                                                =        (        b        −        a        +        1        )        (        c        −        a        )        −                  (                      ∑                          s              =              0                                      b              −              a                                s          )                                                                =        (        b        −        a        +        1        )        (        c        −        a        )        −                              (            b            −            a            )            (            b            −            a            +            1            )                    2                                                                =        (        b        −        a        +        1        )        (        c        −        a        −                              b            2                          +                              a            2                          )                                                        =        (        b        −        a        +        1        )        (        c        −                              b            2                          −                              a            2                          )            for each term. For example,      ∑          i      =      0        6    (  6  −  i  )  =  6  +  5  +  4  +  3  +  2  +  1  =      21    =  7  ⋅  3  =  (  6  −  0  +  1  )  (  6  −            6      2        −            0      2        )    ✓

How to find the values of these kind of summations: <mstyle mathsize="1.2em"> <munderover>

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