1. I hv no other suggestn but do 15.3*2=30.6 2. 48/3=16=3^x, then square to get 9^x=256 3. This is 't i did--1+2+4+...+2^n=2^(n+1)-1, ergo 10(1+2+4+8+16+32)=630, and add the 5 in the begin'g 635 4. I don't thk my way is a shortcut cuz I'm horrible at this kinda stuff, but anyway: multiply by the conjugate 1/(4+3i), so the denominator will be rationalzd to 25, the modulus squared. Then do (5-10i)(4+3i), and merely concerng the integer part you obtain 20+30=50, which, put over the denom 25, yields 2---hmm, obviously thr's sth I don't know about this. 5. nC(m-1)+nCm=(n+1)Cm, so 7C1+7C2=8C2 6. 54*59+55*58=(54*58+54)+55*58=(54+55)58+(58-4)=110*58-4=6380-4=6376. This sounds like an approx problem; and I hope it is. The manipulation above came easily when sitg leisurely, but if it's during a NS test, it maybe a lot harder to derive that on the spot 7. If 1st term is a, 2nd term b, then sum=13a+20b. Where did these qs come from? Best Wishes
if (5-10i)/(4-3i) is a+bi, then a= I dislike these :P In short, the method is:
multiply the top by the bottom's conjugate and remember what a is (don't calculate b) divide by the modulus squared of the bottom (4^2 + 3^2 = 25) a = 20 + 30 = 50 50/25 = 2
7C1 = 7, 7C2 = 21 so 8Cr = 28 r = 2 (8C1 = 8... 8C3 would be too big)
now, a property of combinations is nCr = nC(n-r)
so the bigger choice for N is 8-2 = 6
54*59 + 55*58 - was this approximation?
you could do... xy + (x+1)(y-1) = 2xy + y - x but eh.... I've never seen that before
sum of first 7 = 13x + 20y , x,y = 1 sum = 33 In the real fibonacci sequence (the one that starts with 1 and 1) S(n) is equal to F(n+2) - 1
