Question #85f5f

3 Answers

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

Feb 9, 2018

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

Feb 9, 2018

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 BTony B