how to make sum of (for loop)

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))

[Cedric]

What have you done/tried so far?

[Mech]

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

Answer 1

[Roger Stafford]

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);

Answer 2

[Mech]

help guys :(

[Cedric]

What doesn't work in the solution proposed by Roger?

Answer 3

[David Sanchez]

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;