Răspuns:
Generalizam tipul de suma intalnita in exercitiul 20.
S = 1/1*2 + 1/2*3 + 1/3*4 + ... + 1/(n-1)(n)
1/ab =(b - a)/ab = b/ab - a/ab =
= 1/a - 1/b
Unde b = a+1.
S = 1/1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 ...... + 1/(n-1) - 1/n
S = 1 - 1/n = n/n - 1/n = (n-1)/n
20.
a) rad[102/99 * (50/ 51 - 101/102 + 50/51)] =
= rad[102/99 * (200/202 - 101/102)] =
= rad(102/99 * 99/102) =
= rad(1) = 1
b) rad[128*126*(63/64 - 62/63)] =
= rad[128*126*(63*63 - 62*64)/63*64] =
= rad[4*(63^2 - (63-1)(63+1))] =
= 2rad(63^2 - 63^2 + 1) =
= 2rad(1) =
= 2
c) rad[120/29 * (44/45 - 89/90 + 44/45)] =
= rad[120/29 * (88/45 - 89/90)] =
= rad[120/29 * (176 - 89)/90] =
= rad(120/29 * 87/90) =
= rad(36/9) =
= rad(4) =
= 2.
