How do you find the sum of the terms of geometric 1, -2, 4, -8..... -8192?

1 Answer
George C.
Jun 5, 2016

#-5461#

Explanation:

The general term of a geometric sequence is given by the formula:

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

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

In our example #a = 1# and #r = -2#

If we want to sum the first #N# terms, we find:

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

#=a(r^N-1)#

Dividing both ends by #(r-1)# we find:

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

That's a nice general formula, but note also that:

#sum_(n=1)^N a_n = (a(r^N-1))/(r-1) = (ar^N-a)/(r-1) = (a_(N+1)-a_1)/(r-1)#

So rather than having to count the terms between #1# and #-8192#, we can just use the next term in the sequence, #-8192*-2 = 16364#.

Then:

#sum_(n=1)^N a_n = (a_(N+1)-a_1)/(r-1) = (16384-1)/(-2-1) = -16383/3 = -5461#

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