一條Pure Maths (Summation)

Radius of the ith block = ir(1/n) = ir/n

So, volume of th ith block = pi(ir/n)2(h/n) = pii2r2h/n3

Vn = summation (i = 1→n) pii2r2h/n3

= pir2h / n3 summation (i = 1→ n) i2

= pir2h / n3 [(n)(n + 1)(2n + 1) / 6]

= 1/6 pir2h(1 + 1/n)(2 + 1/n)



Taking limit, as n → infinity, 1/n → 0

So, lim (n → inf.) Vn = 1/6 pir2h(1 + 0)(2 + 0) = 1/3 pir2h