How to evaluate the following sum? 100 100 ⋅ 1 + 100 ⋅ 99 100 2 ⋅ 2 + 100 ⋅ 99 ⋅ 98 100 3 ⋅ 3 + 100 ⋅ 99 ⋅ 98 ⋅ 97 100 4 ⋅ 4 + … + 100 ⋅ 99 ⋅ 98 ⋅ … ⋅ 1 100 100 What I have tried:Above sum can be written as S = ∑ i = 1 100 100 ⋅ 99 ⋅ . . . ⋅ ( 100 − i + 1 ) 100 i ⋅ i = ∑ i = 1 100 100 P i 100 i ⋅ i

Question

How to evaluate the following sum?            100      100        ⋅    1    +                  100        ⋅        99                    100        2              ⋅    2    +                  100        ⋅        99        ⋅        98                    100        3              ⋅    3    +                  100        ⋅        99        ⋅        98        ⋅        97                    100        4              ⋅    4    +    …    +                  100        ⋅        99        ⋅        98        ⋅        …        ⋅        1                    100                  100                     What I have tried:Above sum can be written as       S    =          ∑              i        =        1                    100                            100        ⋅        99        ⋅        .        .        .        ⋅        (        100        −        i        +        1        )                    100        i              ⋅    i        =          ∑              i        =        1                    100                                                            100                                    P          i                            100        i              ⋅    i

perdrigatil5 · Accepted Answer

The sum can be written as      ∑          i      =      0              100                  100      !      i              (      100      −      i      )      !      ⋅              100        i              =  100  !      ∑          i      =      0              100                  100      −      i              i      !      ⋅              100                  100          −          i                      =            100      !              100              100                  ∑          i      =      0              100                  (      100      −      i      )      ⋅              100        i                    i      !        =            100      !              100              100                  ∑          i      =      0              100            (                  100                  i          +          1                            i        !              −                  100        i                    (        i        −        1        )        !              )  which telescopes. Going to the second line from the first, I&#039;ve substituted 100−i for i.Edit in response to OP&#039;s commentsThe substitution does not, in fact, change anything else. The notation       ∑          i      =      0              100        f  (  i  ) simply tells us to sum the terms over the natural numbers ranging from 0 to 100. When you consider       ∑          i      =      0              100        f  (  100  −  i  ), you can see that it gives you the same set of terms, (i.e.   f  (  100  )  ,  f  (  99  )  ,  ⋯  ,  f  (  0  )). And the sigma notation is used as &quot;counting up&quot;, so we do not change the limits.This is indeed neither an arithmetic series nor a geometric series. Are you familiar with telescoping sums? Notice that the expression inside the summation in the final line has the form   f  (  i  +  1  )  −  f  (  i  ). Let us define   f  (  i  )  =            100      i              (      i      −      1      )      !      So we have to evaluate      ∑          i      =      0              100            (    f    (    i    +    1    )    −    f    (    i    )    )    =  f  (  1  )  −  f  (  0  )  +  f  (  2  )  −  f  (  1  )  +  f  (  3  )  −  f  (  2  )  +  ⋯  +  f  (  100  )  −  f  (  99  )  +  f  (  101  )  −  f  (  100  )After cancelling, we have left   f  (  101  )  −  f  (  0  ). Now recall  f  (  i  )  =            100      i              (      i      −      1      )      !        =                    100        i            ⋅      i              i      !      Then   f  (  101  )  =                              100                      101                          ⋅        101                    101        !             and   f  (  0  )  =  0Thus, the sum is            100      !              100              100              ×                    100                  101                    ⋅      101              101      !        =                              100                    Also note that we can change from       ∑    1          100       to       ∑    0          100       because the term corresponding to i=0 is just 0 and does not change the sum.

How to evaluate the following sum (100)/(100)1+(10099)/(100^2)2+ (1009998)/(100^3)3+ (100999897)/(100^4)4+…+(1009998…)/(1100^100)

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