2- suppose a ball is thrown upward to a height of h0 meters . each time the ball bounces, it rebounds to a fraction r of its previous height . let hn be the hight after the nth term biunces. cosider the following value of h0 and r
a- find the first 4 terms of the sequence of heights(hn)
b- find a general expression for the nth term of the sequence (hn)
h0=20 , r=0.5

a) 40, 50, 55, 57.5

b)

Step-by-step explanation:

h₀ = Initial height of the ball =20

r = Rebound fraction = 0.5

a) The series of bouncing balls is given by

Sₙ=h₀+2h₀(r¹+r²+r³+r⁴.........rⁿ)

S₁=h₀+2h₀r¹=20+2×20×0.5=40

S₂=h₀+2h₀(r¹+r²)=20+2×20×(0.5+0.5²)=50

S₃=h₀+2h₀(r¹+r²+r³)=20+2×20×(0.5+0.5²+0.5³)=55

S₄=h₀+2h₀(r¹+r²+r³+r⁴)=20+2×20×(0.5+0.5²+0.5³+0.5⁴)=57.5

b) General expression for the nth term of the sequence

step-by-step explanation:

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step-by-step explanation: 4 cause 2+2=4