How do you find the sum of the first six terms of the geometric sequence 1, 1.5, 2, 2.25, ...?

1 Answer
George C.
Dec 9, 2015

This is not a geometric sequence. The term #2# should not be there.

Without the #2#, the first #6# terms of the geometric sequence sum to #20.78125#

Explanation:

Discarding the errant term #2#, the sequence #1#, #1.5#, #2.25# is a geometric sequence with common ratio #1.5#.

The general formula for a term of a geometric sequence is:

#a_n = a r^(n-1)#

where #a# is the initial term and #r# is the common ratio.

For our sequence #a=1# and #r = 1.5#

Then:

#(r-1) sum_(n=1)^N a_n#

#=(r-1) sum_(n=1)^N a r^(n-1)#

#=r sum_(n=1)^N a r^(n-1) - sum_(n=1)^N a r^(n-1)#

#=sum_(n=2)^(N+1) a r^(n-1) - sum_(n=1)^N a r^(n-1)#

#= a r^N + color(red)(cancel(color(black)(sum_(n=2)^N a r^(n-1)))) - a - color(red)(cancel(color(black)(sum_(n=2)^N ar^(n-1))))#

#= a(r^N-1)#

So:

#sum_(n=1)^N a_n = (a(r^N-1))/(r-1)#

So for our sequence:

#sum_(n=1)^6 a_n = (1(1.5^6-1))/(1.5-1)#

#=2((3/2)^6-1)#

#=2(729/64-1)#

#=(729-64)/32#

#=665/32#

#=20.78125#

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