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generate a summation series

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Odien
Odien il 13 Ago 2015
Modificato: Walter Roberson il 12 Giu 2021
3(2+1)+4(3+2+1)+5(4+3+2+1)+6(5+...1)+...+1000(999+...+1).
How to write a matlab code to calculate the following summation without using for loop?
i can see the trend for 3(2+1) if the n = 1 (n+2)[((n+2)-1)+((n+2)-2)].
Assume the summation only sum until 6th term.
can we use array to solve this?

Risposta accettata

Stephen23
Stephen23 il 13 Ago 2015
Modificato: Stephen23 il 13 Ago 2015
>> X = 1+cumsum(2:999); % 1+[2,3+2,4+3+2,...,999+..+2]
>> Y = 3:1000; % [3,4,5,...,1000]
>> sum(X.*Y)
ans = 125083208248
Note that this uses element-wise multiplication.
  2 Commenti
Odien
Odien il 13 Ago 2015
Thank you !
Walter Roberson
Walter Roberson il 13 Ago 2015
This looks like it was homework to me...

Accedi per commentare.

Più risposte (1)

Walter Roberson
Walter Roberson il 13 Ago 2015
It does have a symbolic answer that you could find using nested symsum() if you have the symbolic toolbox.
But for a numeric answer: cumsum() and multiply by something gives a bunch of terms...

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