Algebra 2: Sequences and Series

E

edd fedds

New member
Joined
Dec 11, 2009
Messages
26
* NeedHelp on these Problem. I solved the poblems but came out with wrong solutions being that they do not match the answer choices. Please, need Help! Thank you. My questions and solutions are below. :mrgreen:

-What is Sn of the geometric series with a1 = 4, an = 256 and n = 4?
a. 84
b. 260
c. 336
d. 340
*My solution was:
Sn=n(a[1]+a[n])/2
=4(4+256)/2
=4(260)/2=520. Need Help...

-What is Sn of the geometric series with a1 = -81, r = -2/3, and n = 5?
a. -55
b. -39
c. 26
d. 211
*Sn=a[1]-a[1]r^n/1-r
=(-81)-(-81)-2/3^5/1-(-2)
=(-81)-(-81)(-32/243)/3
=(-275/3)/3
=-275/9
=-30.55.???

-What the sum of the geometric sequence ?
a. 240
b. 363
c. 726
d. 9330
*My solution was:
5 (geometric sequence symbol) 2(3)^n
=2(3)^1+2(3)^2+2(3)^3+2(3)^4+2(3)^5
=6+18+54+162+486.
Then, a[1]=6, a[5]=486,r=6,
Sn= a[1]-a[n]^r/1-r
=6-486(6)/1-(6)
=6-2916/-5=582.

-What is the sum of the infinite geometric series with a1 = 27 and r = -4/5?
a. 11
b. 15
c. 135
d. does not exist
*My solution was:
Sn=a[1]-a[1]r^n/1-r
=27-27(-4/5)^1/1-(-45)
=27-27(-4/5)/a/5=???

-What is the sum of the infinite geometric series with a1 = 42 and r = 6/5?
a. 105
b. 19
c. -210
d. does not exist
*How do I solve?


-What is the sum of the infinite geometric series 1/3 + 1/9 + 1/27 + . . . ?
a. 1/2
b. 3/4
c. 1
d. does not exist
*I chose choice B. Am I right?


What is the fifth term of the binomial expansion (x + y)7?
a. 21x5 y2
b. 35x3 y4
c. 35x4 y3
d. 21x2 y5
*I chose choice C. Am I right?

Tak you so much again!
 
edd fedds said:
* NeedHelp on these Problem. I solved the poblems but came out with wrong solutions being that they do not match the answer choices. Please, need Help! Thank you. My questions and solutions are below. :mrgreen:

-What is Sn of the geometric series with a1 = 4, an = 256 and n = 4?
a. 84
b. 260
c. 336
d. 340
*My solution was:
Sn=n(a[1]+a[n])/2 <<< This is expression for sum of Arithmatic Series
=4(4+256)/2
=4(260)/2=520. Need Help...

-What is Sn of the geometric series with a1 = -81, r = -2/3, and n = 5?
a. -55
b. -39
c. 26
d. 211
*Sn=a[1]-a[1]r^n/1-r
=(-81)-(-81)-2/3^5/1-(-2)
=(-81)-(-81)(-32/243)/3
=(-275/3)/3
=-275/9
=-30.55.???

-What the sum of the geometric sequence ?
a. 240
b. 363
c. 726
d. 9330
*My solution was:
5 (geometric sequence symbol) 2(3)^n
=2(3)^1+2(3)^2+2(3)^3+2(3)^4+2(3)^5
=6+18+54+162+486.
Then, a[1]=6, a[5]=486,r=6,
Sn= a[1]-a[n]^r/1-r
=6-486(6)/1-(6)
=6-2916/-5=582.

-What is the sum of the infinite geometric series with a1 = 27 and r = -4/5?
a. 11
b. 15
c. 135
d. does not exist
*My solution was:
Sn=a[1]-a[1]r^n/1-r
=27-27(-4/5)^1/1-(-45)
=27-27(-4/5)/a/5=???

-What is the sum of the infinite geometric series with a1 = 42 and r = 6/5?
a. 105
b. 19
c. -210
d. does not exist
*How do I solve? <<< Use equation given below


-What is the sum of the infinite geometric series 1/3 + 1/9 + 1/27 + . . . ?
a. 1/2
b. 3/4
c. 1
d. does not exist
*I chose choice B. Am I right? <<<< NO - show work so that we can correct it


What is the fifth term of the binomial expansion (x + y)7?
a. 21x5 y2
b. 35x3 y4
c. 35x4 y3
d. 21x2 y5
*I chose choice C. Am I right?

Tak you so much again!
Click to expand...

Sum of geometric series:

\(\displaystyle S_n \ = \ a_1 \ + \ a_1 * r \ + \ a_1*r^2 \ + \ a_1*r^3 \ + \ a_1*r^4 \ + \ ... \ + \ a_1*r^{n-1}\)

\(\displaystyle S_n \ = \ a_1 \cdot \frac{r^n \ - \ 1}{r \ - \ 1}\)
 
edd fedds said:
-What is Sn of the geometric series with a1 = -81, r = -2/3, and n = 5?
a. -55
b. -39
c. 26
d. 211

Sn = (a[1] - a[1] r^n)/(1 - r)

Grouping symbols are VERY important, when typing algebraic ratios. Without them, the typing is garbage. (It also helps readability, when word spaces are placed around subtraction and addition signs, like above.)

S[n] = (-81)-(-81)-2/3^5/1-(-2) Hmmm, this is way off.

It should be:

S[n] = [(-81) - (-81)(-2/3)^5]/[1 - (-2/3)]

We don't subtract 2/3^5, and how did r change from -2/3 to -2? If you can't locate errors like these yourself, you've got rough waters ahead, mate!
Click to expand...

How about we use the formula for S[n] provided by Subhotosh; your version is not "standard".

S[n] = a[1] * (r^n - 1)/(r - 1)

S[n] = (-81) * ([-2/3]^5 - 1)/(-2/3 - 1)

Actually, you can continue with your version corrected by me above, if you prefer. I don't care. 8-)

Thank you for showing your work; it makes it clear to us what went wrong.
 
Oh my gosh! Thank you so much guys. I can't believe that I looked over that when I was solving that equation. Thanks for pointing that out to me mmm4444bot. I need to slow down a bit. :lol: Thanks though!
 
Reduce your speed by half and get twice as much done :wink:
 
You must log in or register to reply here.
Top