Lim_ {n} in {}} {_} {n} {f} {n} [(frac {i} {n}) ^ 2 + 1]. ... ??

Odgovor:

#4#

Pojasnilo:

# = lim_ {n-> oo} (3 / n ^ 3) sum_ {i = 1} ^ {i = n} i ^ 2 + (3 / n) sum_ {i = 1} ^ {i = n} 1 #

# "(Faulhaberjeva formula)" #

# = lim_ {n-> oo} (3 / n ^ 3) (n (n + 1) (2n + 1)) / 6 + (3 / n) n #

# = lim_ {n-> oo} (3 / n ^ 3) n ^ 3/3 + n ^ 2/2 + n / 6 + (3 / n) n #

# = lim_ {n-> oo} 1 + ((3/2)) / n + ((1/2)) / n ^ 2 + 3 #

# = lim_ {n-> oo} 1 + 0 + 0 + 3 #

#= 4#

Odgovor:

# 4#.

Pojasnilo:

Tukaj je drugo pot rešiti Težava:

Spomnimo se, da # int_0 ^ 1f (x) dx = lim_ (n do oo) sum_ (i = 1) ^ n1 / nf (i / n) … (zvezda) #.

#:. "Reqd. Lim. =" Lim_ (n do oo) sum_ (i = 1) ^ n3 / n {(i / n) ^ 2 + 1} #, # = 3 lim_ (n do oo) vsota_ (i = 1) ^ n1 / n {(i / n) ^ 2 + 1} #, # = 3int_0 ^ 1 {(x) ^ 2 + 1} dx ………… ker, (zvezda) #,

# = 3 x ^ 3/3 + x _0 ^ 1 #, # = x ^ 3 + 3x _0 ^ 1 #, # = 1 ^ 3 + 3xx1- (0 ^ 3 + 3xx0) #, #rArr "Reqd. Lim. =" 4 #.