= β ( 1 (1 + 100 β ) β1+ ) =1 + ( 100 (1 + 100 β ) β1+ ) =
π·π =π
πβ
(
1
(1 +π¦
100 β π)πβ1+π‘
)
π
π=1
+
(
100
(1 +π¦
100 β π)πβ1+π‘
)
π‘ =π·ππ
π·πΆπΆ