Ok either the book is wrong or I just found out that I fail at basic arithmetic sequences... The book says -1845. I say 'huh'?
When your done here is my answer: Spoiler 80 Assuming we agree I would say it makes a better case for the book being wrong.
Well I assume you are correct since you got higher mathematic education than me however my answer does not match yours: S10=5(2+36) . S10=5(18-27) ==> 190x-45 = -8550 I want my money back for basic education.
Is it? Do you care to show me your working? Since everyone I asked seems to fail at everything when it comes to solving this question so I would love to put an end to this madness.
How are you solving the question by hand or with a calculator? The way I would do it is put sum((-3 4i)(12-3i),i,1,10) into a calculator and let it plug and chug (which produces the book's answer). I actually multiplied it out and created an error which I caught because my answer just seemed to small given how negative the function is suppose to get. If your doing it by hand than good luck. The way that I think would create the least number of errors is write out the values for each function (for i from 1 to 10) seperately then multiply them together one by one and then add all of the results together.
Which would look like this: i ui vi ui*vi 1 1 9 9 2 5 6 30 3 9 3 27 4 13 0 0 5 17 -3 -51 6 21 -6 -121 7 25 -9 -225 8 29 -12 -348 9 33 -15 -495 10 37 -18 -666 all added together -1845
Yep just finished it and got the correct answer. Blargh I must be doing something wrong. Shouldn't finding sum for u1 and v1 and then multiplying the outcomes be sufficient? As in Eu1 . Ev1
No, that doesn't work because the result of a multiplication grows faster than the result of an addition as the inputs grow, which is why there is an order of operations. Just like you have to do the multiplication in 3*4+1*2 (which equals 14) or you get a bad result (if you do the addition first you get 30) you have to do the multiplcation in a sumation before the sumation itself (the sumation operation is on the same operation level as addition). Basicly, sum(x+2x,x,1,10) = sum(x,x,1,10) + sum(2x,x,1,10) but sum(x*2x) != sum(x,x,1,10)*sum(2x,x,1,10).
Yes that makes perfect sence. At first I did think the same thing however my teacher persuaded me that Eui . Evi should work. Since he nor the rest of the group could do it I assumed there is something more to it but apparently I was just wrong (and the rest of us). I'll have to do quick revision on multiplication of arithmetic sequences and get back on my teacher. Thanks.
To reiterate, SUM{ u_i * v_i } is NOT the same as SUM{u_i } * SUM{v_i} but, SUM{ a_i + b_i } IS the same as SUM{a_i } + SUM{b_i} So what one can do is expand the summation. u_i * v_i = (-3+4i)*(12-3i) = (-36 + 48i + 9i - 12i^2) = (-12i^2 + 57i - 36) SUM{u_i*v_i} = SUM{12i^2 + 57i - 36} = SUM{-12i^2} + SUM{57i} - SUM{36} = -12.SUM{i^2} + 57.SUM{i} - 36.SUM{1} = -12.{n(n+1)(2n+1)/6} + 57.{n(n+1)/2} - 36.{n} ; where n=10 = -2n(n+1)(2n+1) + 57n(n+1)/2 - 36n we could insert n=10 there, or we could tidy it up a little more first, = (n/2).( -4(n+1)(2n+1) + 57(n+1) - 72 ) = (n/2).(-8n^2 - 12n - 4 + 57n + 57 - 72) = (n/2).(-8n^2 + 45n - 19) letting n=10 now; = (5)* (-8*100 + 45*10 - 19) = 5 * (-800 + 450 - 19) = 5 * (-369) = -1845
