# how to make sum of (for loop)

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Mech on 7 Jun 2014
Answered: David Sanchez on 9 Jun 2014
I need for loop or any method to calculat next sum
• m=1:n
• a+2(cos(b))
• a+2(cos(b)+cos(2b))
• a+2(cos(b)+cos(2b)+cos(3b))
• a+2(cos(b)+cos(2b)+.............cos(nb))
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Mech on 7 Jun 2014
Edited: Mech on 7 Jun 2014
ok
• a1=a+2(cos(b))
• a2=a+2(cos(b)+cos(2b))
• a3=a+2(cos(b)+cos(2b)+cos(3b))
I need value of a1,a2,a3

Roger Stafford on 7 Jun 2014
Edited: Roger Stafford on 7 Jun 2014
Assuming b is a scalar,
s = a + 2*sum(cos((1:n)*b));
An alternate formula without the long summation is:
s = a + 2*cos((n+1)*b/2)*sin(n*b/2)/sin(b/2);

Mech on 8 Jun 2014
help guys :(
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Cedric Wannaz on 9 Jun 2014
What doesn't work in the solution proposed by Roger?

David Sanchez on 9 Jun 2014
the for loop:
a = 3; % or whatever value you have in mind
b = 2; % or whatever value you have in mind
n = 10; % or whatever value you have in mind
your_sum = 0; % initialization of summation part
for m=1:n
your_sum = your_sum + cos(m*b);
end
your_sum = a + 2*your_sum;