What is the 7th partial sum of #sum_(i=1)^∞ 6(3)^(i-1)#?

1 Answer
Steve M
Dec 4, 2016

6558

Explanation:

# \ \ \ \ \ sum_(i=1)^oo 6(3)^(i-1) = sum_(i=1)^oo (2*3)(3)^(i-1)#
# :. sum_(i=1)^oo 6(3)^(i-1) = sum_(i=1)^oo (2)(3)^i#
# :. sum_(i=1)^oo 6(3)^(i-1) = 2 sum_(i=1)^oo (3)^i#
# :. sum_(i=1)^oo 6(3)^(i-1) = 2 {3^1+3^2+3^3+ 3^4+...}#
# :. sum_(i=1)^oo 6(3)^(i-1) = 2 {3+3*3+3*3^2+ 3*3^3+...}#

And the expression in #{...}# is a GP with #a=r=3# and so we can use the GP sum formula #S_n = a(1-r^n)/((1-r)# to get:

# \ \ \ \ \ sum_(i=1)^oo 6(3)^(i-1) = 2 *3(1-3^n)/(1-3)#
# :. sum_(i=1)^oo 6(3)^(i-1) = 6(3^n-1)/2#
# :. sum_(i=1)^oo 6(3)^(i-1) = 3(3^n-1)#

And so the required sum is:
# sum_(i=1)^7 6(3)^(i-1) = 3(3^7-1) = 6558#

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