F = t ⇨ df = dt
dg = sec² 2t dt ⇨ g = (1/2) tan 2t
⇔
integral of t sec² 2t dt = (1/2) t tan 2t - (1/2) integral of tan 2t dt
u = 2t ⇨ du = 2 dt
As integral of tan u = - ln (cos (u)), you get :
integral of t sec² 2t dt = (1/4) ln (cos (u)) + (1/2) t tan 2t + constant
integral of t sec² 2t dt = (1/2) t tan 2t + (1/4) ln (cos (2t)) + constant
integral of t sec² 2t dt = (1/4) (2t tan 2t + ln (cos (2t))) + constant ⇦ answer
Since it’s a square we know that GJ has to be the same as GH. To find a diagonal you would do a^2 + b^2 = c^2. Which gives us about 9.19. So GJ=6.5 and JH=9.2
Find the formula for the polynomial
then solve
Answer:
an = (1 + 2·(n - 1))/2^(n - 1) = 2^(1 - n)·(2·n - 1)
sn = 6 - 2·0.5^n·(2·n + 3)