Let k be a integer. How can we compute the close formula for ∑ m = 0 k ( m + 1 ) ( m + 2 ) ( 2 m + 3 ) ( 3 m + 4 ) ( 3 m + 5 ) ?

Question

Let k be a integer. How can we compute the close formula for      ∑          m      =      0              k        (  m  +  1  )  (  m  +  2  )  (  2  m  +  3  )  (  3  m  +  4  )  (  3  m  +  5  )  ?

Myla Pierce · Accepted Answer

We can expand the polynomial we are summing over to give a degree 5 polynomial in k, as follows:  (  m  +  1  )  (  m  +  2  )  (  2  m  +  3  )  (  3  m  +  4  )  (  3  m  +  5  )  =  18      m          5        +  135      m          4        +  400      m          3        +  585      m          2        +  422  m  +  120Therefore, we can write the summation as:  18      ∑          m      =      0              k                  m              5              +  135      ∑          m      =      0              k                  m              4              +  400      ∑          m      =      0              k                  m              3              +  585      ∑          m      =      0              k                  m              2              +  422      ∑          m      =      0              k            m    +  120      ∑          m      =      0              k        1We can therefore compute the final answer without much difficulty using standard results:      ∑          m      =      0              k                  m              5              =                    k                  2                    (      2              k                  2                    +      2      k      −      1      )      (      k      +      1              )                  2                      12        ∑          m      =      0              k                  m              4              =                  k        (        2        k        +        1        )        (        k        +        1        )        (        3                  k                      2                          +        3        k        −        1        )            30            ∑          m      =      0              k                  m              3              =                    k                  2                    (      k      +      1              )                  2                      4        ∑          m      =      0              k                  m              2              =            k      (      k      +      1      )      (      2      k      +      1      )        6        ∑          m      =      0              k            m    =            k      (      k      +      1      )        2            ∑          m      =      0              k            1    =  k  +  1Combining all these, we have the following simplified expression:  (  k  +  2      )          2            (    3          k              4              +    24          k              3              +    67          k              2              +    76    k    +    30    )  Testing for a few simple test cases:                    k        =        1        :        1800                            k        =        2        :        11040                            k        =        4        :        133560            These correspond to what we get if we compute the summation by hand, so this is our final closed form.

Let k be a integer. How can we compute the close formula for <munderover> ∑

Answered question

Answer & Explanation

New Questions in Calculus 2