How to evaluate the following sum? 100 100 ⋅ 1 + 100 ⋅ 99 100...

Kaitlin Jacobson

Kaitlin Jacobson

Answered

2022-12-19

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

Answer & Explanation

perdrigatil5

perdrigatil5

Expert

2022-12-20Added 2 answers

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've substituted 100−i for i.
Edit in response to OP's comments
The 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 "counting up", 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 ) = 0
Thus, 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.

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