oh wait nvm i see what i did wrong
shouldn't it be 1 * (x-3)^2 + 2x(x-3)^1 ? instead of 2x(x-3) * 1 at the end?
differentiate x(x-3)^2 do I product and chain rule?
r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n r(1+i)^-n [ (1+i)^(n-1) + (1+i)^(n-2) + .. + (1+i)^2 + (1+i)^1 + 1] how did you get the (1+i)^2? and why isnt there a (1+i)^ n+3?
oh yea i did forget the r in my first sum thank you reiny i think i get it now
im assuming you replaced (1+i) with u right if so i thinki kinda get
1) equation; sum= (1-(1+i)^-n / i 2) equation; sum= r(1+i)^-1 + r(1+i) ^-2 + r(1+i)^3 +...+ r(1+i)^-(n-1) + r(1+i)^-n im suppose to simplify second equation into the first one but im messing up somewhere if anyone can tell me the first 3-4 steps i think i can get it from there...
stop spamming the same question like 10 times
im not going to do it but ill give you a hint you need common denominators to add or subtract fraction so figure out what you need to do to both fractions to get same denominator
Math: Check Answers
i didnt check number 4 but the others should be right