What is the sum of the first 6 terms of this geometric sequence? 1,8,64,512

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Tony B Share
Feb 10, 2018

Answer:

#37449#

ALL background information given

Explanation:

#color(blue)("Understanding how the sequence works")#

Let the term count be #i#
Let the #i^("th")# term be #a_i#

Then we have:

#i=1->a_1=1#
#i=2->a_2=8#
#i=3->a_3=64#
#i=4->a_4=512#

The rate of increase is far to great for an arithmetic progression so it is a geometric sequence as stated in the question.

Lets have a play!

From our multiplication tables notice that #8xx8=64# so this could be a link.

Lets try #8^3=512# which works. So we have something involving #xx8#

At a guess lets investigate:

#a_1=1color(green)(larr" I will come back to this")#
#a_2=8=1xx8#
#a_3=64=1xx8xx8#
#a_4=512=1xx8xx8xx8#

So by observation, what is happening is that we have for any term where #i>1color(white)("dd")# #a_i=1xx8^(i-1)#

Given that #a_1=1#
#a_i=a_2=8=a_1xx8^(2-1)=8^1#
#a_i=a_3=64=a_1xx8^(3-1)=8^2#
#a_i=a_4=512=a_1xx8^(4-1)=8^3#

How does #a_1# fit this process. Note that #8^0=1# so for
#a_i=a_1=1xx8^(1-1)color(white)("dd")=color(white)("dd")1xx8^0color(white)("dd")=color(white)("dd")1xx1=1#

So it works for #a_1# as well thus we are now are working with #i > 0#

Thus the general rule is #a_i=1xx8^(i-1) = 8^(i-1)#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Determine the sum to any term " a_n)#

Set #s=1+8^1+8^2+8^3+...8^(n-1)" ".......Equation(1)#

Multiply #s# by 8 giving:

#8s=8^1+8^2+8^3+...8^(n-1)+8^n" "......Equation(2)#

Subtract #Eqn(2)" from "Eqn(1)#

#8s-s=8^n-1#

Factor out #s#

#s(8-1)=8^n-1#

#s=(8^n-1)/(7)#

#color(brown)("So summing to the "6^("th")" term we have:")#

#s=(8^6-1)/(7)#

#color(white)("dddddddddd")color(blue)( ul(bar(|color(white)(2/2)s=37449color(white)(2/2)|)))#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#
Tony B

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Jim G. Share
Feb 9, 2018

Answer:

#S_(6)=37449#

Explanation:

#"using the sum to n terms for a geometric sequence"#

#•color(white)(x)S_n=(a(1-r^n))/(1-r)=(a(r^n-1))/(r-1)#

#"where a is the first term and r the common ratio"#

#r=a_2/a_1=a_3/a_2= ...... =a_n/a_(n-1)#

#rArrr=8/1=64/8=512/64=8" and "a=1#

#rArrS_6=(1(8^6-1))/7=37449#

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Feb 9, 2018

Answer:

#sum_(k=1)^6 (1)8^(k-1)= 37449#

Explanation:

From the reference Geometric Progression we obtain the equation:

#a_n = ar^(n-1)#

We can find the value of a; given that #a_1 = 1#

#a_1 = ar^(1-1)#

#1 =ar^0#

#a = 1#

We can find the value of r; given that #a_2 = 8# and knowing that #a = 1#

#a_2 = (1)r^(2-1)#

#r = 8#

The same reference gives us the equation:

#sum_(k=1)^n ar^(k-1)= (a(1-r^n))/(1-r)#

We know that #a = 1# and #r=8# and we want the first 6 terms which means that #n=6#:

#sum_(k=1)^6 (1)8^(k-1)= (1(1-8^6))/(1-8)#

#sum_(k=1)^6 (1)8^(k-1)= 37449#

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