How do you find a formula for the sum n terms #Sigma (1+(2i)/n)^3(2/n)# and then find the limit as #n->oo#?

1 Answer
Nov 15, 2016

Answer:

# sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 2/n^2(10n^2+13n+4)#
# lim_(n rarr oo)sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 20#

Explanation:

We will need the standard results:
# sum_(r=1)^n r = 1/2n(n+1) #
# sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) #
# sum_(r=1)^n r^2 = 1/4n^2(n+1)^2 = (sum_(r=1)^n r)^2 #

Let

# S_n = sum_(i=1)^n(1 + (2i)/n)^3(2/n) #
# :. S_n = 2/n sum_(i=1)^n((n + 2i)/n)^3 #
# :. S_n = 2/n sum_(i=1)^n1/n^3(n + 2i)^3 #
# :. S_n = 2/n^4 sum_(i=1)^n(n + 2i)^3 #
# :. S_n = 2/n^4 sum_(i=1)^n(n^3 + 3n^2(2i) + 3n(2i)^2 + (2i)^3) #
# :. S_n = 2/n^4 sum_(i=1)^n(n^3 + 6n^2i + 12ni^2 + 8i^3) #
# :. S_n = 2/n^4 {sum_(i=1)^n n^3 + sum_(i=1)^n 6n^2i + sum_(i=1)^n 12ni^2 + sum_(i=1)^n 8i^3 }#
# :. S_n = 2/n^4 {n^3sum_(i=1)^n 1 + 6n^2sum_(i=1)^n i + 12nsum_(i=1)^n i^2 + 8sum_(i=1)^n i^3 }#
# :. S_n = 2/n^4 {n^3 n + 6n^2 1/2n(n+1) + 12n 1/6n(n+1)(2n+1) + 8 1/4n^2(n+1)^2 }#
# :. S_n = 2/n^4 {n^4 + 3n^3 (n+1) + 2n^2(n+1)(2n+1) + 2n^2(n+1)^2 }#
# :. S_n = 2/n^4 n^2 {n^2 +3n (n+1) + 2(n+1)(2n+1) + 2(n+1)^2 }#
# :. S_n = 2/n^2 {n^2 +3n^2+3n + 2(2n^2+3n+1) + 2(n^2+2n+1) }#
# :. S_n = 2/n^2{n^2 +3n^2+3n + 4n^2+6n+2 + 2n^2+4n+2 }#
# :. S_n = 2/n^2(10n^2+13n+4)#

Hence,

# sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 2/n^2(10n^2+13n+4)#

And so;

# lim_(n rarr oo)S_n = 2/n^2(10n^2+13n+4)#
# lim_(n rarr oo)S_n = 20 + 26/n + 8/n^2 #
# lim_(n rarr oo)S_n = 20 # as both #1/n rarr 0#, and #1/n^2 rarr 0# as #n rarr oo#

Hence,

# lim_(n rarr oo)sum_(i=1)^n(1 + (2i)/n)^3(2/n)= 20#