f : R \rightarrow R, f(x) = 5x - 1
S = f(0) - f(1) + f(2) - f(3) + ... + f(50)

I think you might have made a typo don't you mean f(0)+f(1)+f(2) etc?

Algebra isn't that hard all you have to do for every f(#) is replace the x by the #. For example f(x)=5x-1 ->
f(0)=5*0-1=-1
f(1)=5*1-1=5-1=4

Now go do your homework, lol

I think you might have made a typo don't you mean f(0)+f(1)+f(2) etc?


Algebra isn't that hard all you have to do for every f(#) is replace the x by the #. For example f(x)=5x-1 ->
f(0)=5*0-1=-1
f(1)=5*1-1=5-1=4

Now go do your homework, lol

No, there are no mistakes in the enuciation of the problem.

If it had been just a sum, I wouldn't have been asked for some tips because it's obvious that the solution would have been S = 5 (1 + 2 + 3 +... + 50) + 51 = 5 (50 * 51) / 2 + 51 = etc. but in this case it looks like S = 5 (-1 + 2 - 3 + 4 -5 +... + 50) + k * 1. So how could I solve for this?

Hint: consider that the sum of f(n) - f(n+1) always equals -5 for any even value of 'n.' Can you take it from here?

Hint: consider that the sum of f(n) - f(n+1) always equals -5 for any even value of 'n.' Can you take it from here?

Thank you a lot !