Lim n approaches infinity# 6/n((2n)/3 + (5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2))=#?
lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?
lim n approaches infinity [6/n(2n/3 + 5n(n+1)/2n - 4n(n+1)(2n+1)/6n^2)]=?
3 Answers
Explanation:
After expanding
so
I'm guessing this is typed incorrectly.
Explanation:
The term 5n(n+1)/2n means
My guess is the question wants (5n(n+1))/(2n) which is
Similarly, 4n(n+1)(2n+1)/6n^2 is
So here's what I guess the expression should be:
# = (4n)/n + (15n(n+1))/n^2 - (4n(n+1)(2n+1))/n^3#
# = 4 +15((n/n)((n+1)/n)) - 8(n/n)((n+1)/n)((2n+1)/n)#
The limit as
#4 +15(1)(1)-8(1)(1)(2) = 4+15-8 = 11#
# lim_(n rarr oo) 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] = 11 #
Explanation:
Let
# S = 6/n[(2n)/3 +( 5n(n+1))/(2n) - (4n(n+1)(2n+1))/(6n^2) ] #
# :. S = 6[ 1/(6n^2){ 4n^2 + 15n(n+1) - 4(n+1)(2n+1) } ] #
# :. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 4(2n^2+3n+1) ] #
# :. S = 1/(n^2)[ 4n^2 + 15n^2+15n - 8n^2-12n-4 ] #
# :. S = ( 11n^2 + 3n - 4 ) / (n^2)#
# :. S = 11 + 3/n - 4/n^2 #
So:
# lim_(n rarr oo) S = lim_(n rarr oo) (11 + 3/n - 4/n^2) #
# lim_(n rarr oo) S = lim_(n rarr oo) 11 + 3lim_(n rarr oo) 1/n - 4lim_(n rarr oo)1/n^2 #
# lim_(n rarr oo) S = 11 + 0 + 0#
# lim_(n rarr oo) S = 11 #
This ca be verified by looking at the graph of
graph{11 + 3/x - 4/x^2 [-27.95, 76.05, -19, 33]}