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1 year ago

Consider the series 4 + 12 + 36 ... Find the sum of the first 9 terms.

Mathematics

1 answer:

jeyben [28]1 year ago

3 0

Answer:

39364

Step-by-step explanation:

You can simplify the equation to make 4(1+3+9...) The nth term is therefore $3^{n-1}$ so u can substitute that into the equation so 4( $3^{n-1}+ 3^{n} + 3^{n+1}$ ..). This would simplify to 4( $3^0 +3+3^2$ + $3^3+3^4+3^5+3^6+3^7+3^8$ ). By simplifying this, you can make it so 4( $2^2+6^2+18^2+42^2+81^2$ ) = 4(4+36+324+2916+6561) =39364

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3 years ago

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Answer:

a) $a_n=5\,\,(2)^{n-1}$

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Step-by-step explanation:

a) Geometric sequence with first term 5 and common ratio 2, where the nth term can be calculated via:

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The first five terms are: $a_1=5;\,\,\,a_2=10;\,\,\,a_3=20; \,\,\,a_4=40;\,\,\,a_5=80$

b) Geometric sequence with first term 100 and common ratio 1/2, where the nth term can be calculated via:

$b_n=100\,\,(\frac{1}{2} )^{n-1}$

The first five terms are: $a_1=100;\,\,\,a_2=50;\,\,\,a_3=25; \,\,\,a_4=12.5;\,\,\,a_5=6.25$

c) Geometric sequence with first term 160 and common ratio -1/2, where the nth term can be calculated via:

$c_n=160\,\,(-\frac{1}{2} )^{n-1}$

The first five terms are: $a_1=160;\,\,\,a_2=-80;\,\,\,a_3=40; \,\,\,a_4=-20;\,\,\,a_5=10$

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2 years ago

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Answer:

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1 year ago

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Step-by-step explanation:

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You haven't provided the steps.

mathisfun.com/geometry/construct-linebisect.html

Here is a useful link to the correct steps. The instructions may not be exactly the same but I think you can do it.

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2 years ago

Consider the series 4 + 12 + 36 ... Find the sum of the first 9 terms.​

Consider the series 4 + 12 + 36 ... Find the sum of the first 9 terms.