Borrar filtros
Borrar filtros

generate a summation series

5 visualizaciones (últimos 30 días)
Odien
Odien el 13 de Ag. de 2015
Editada: Walter Roberson el 12 de Jun. de 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?
  2 comentarios
ibraheem kanai
ibraheem kanai el 5 de Jun. de 2021
Walter Roberson
Walter Roberson el 5 de Jun. de 2021
Editada: Walter Roberson el 12 de Jun. de 2021

Iniciar sesión para comentar.

Respuesta aceptada

Stephen23
Stephen23 el 13 de Ag. de 2015
Editada: Stephen23 el 13 de Ag. de 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 comentarios
Odien
Odien el 13 de Ag. de 2015
Thank you !
Walter Roberson
Walter Roberson el 13 de Ag. de 2015
This looks like it was homework to me...

Iniciar sesión para comentar.

Más respuestas (1)

Walter Roberson
Walter Roberson el 13 de Ag. de 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...

Categorías

Más información sobre Mathematics en Help Center y File Exchange.

Productos

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by