I got the same answer but youre supposed to calculate q and simplify the answers with it which i didnt. How many marks do u think i will lose? (I left the y coordinate and area in terms of a and q)
for the last induction question i changed f(k+1) into the same format as f(k) and then when i added 11f(k) it was divisible by 19. does this sound vaguely right to anyone?? had no idea what i was doing lol
How did you solve this one, tbh I just added it 12 times
So you know the summation formula which was n/2(n+3)(2n+5) if i recall correctly, put 12 in and you got 2610. so 2610 + k(2^(r-1)) = 3520 k(2^(r-1)) = 910 now as r is increasing you need to do K(2^0 + 2^1 + 2^2... 2^11) = 910 it came to 910/4095 = 2/9
So you know the summation formula which was n/2(n+3)(2n+5) if i recall correctly, put 12 in and you got 2610. so 2610 + k(2^(r-1)) = 3520 k(2^(r-1)) = 910 now as r is increasing you need to do K(2^0 + 2^1 + 2^2... 2^11) = 910 it came to 910/4095 = 2/9