11.1 Types of Sequences
pg 545 #1
The sequence 56, 28, 14, 7, ... is ____
arithmetic, geometric, or neither
Geometric
pg 546 #2
Find the formula for the nth term of the arithmetic sequence.
-7, -3, 1, 5,...
t_1=-7
d=4
t_n=-7+4(n-1)
pg 546 #6
Find the geometric mean of 10 and 40
root()((10)(40))=root()(400)=20
pg 546 #7
Write the series in expanded form
\sum_(k=0)^(3)2^(k-1)
t_1=1/2
t_2=1
t_3=2
t_4=4
1/2+1+2+4
pg 546 #9
Find the sum of the arithmetic series
\sum_(n=1)^(25)(3n+2)
n=25,d=3,t_1=5,t_25=77
S_n=(n(t_1+t_n))/2
S_25=(25(5+77))/2=1025
pg546 #11
Find the sum of the infinite geometric series if it has one
64+48+36+27+...
r=48/64=3/4
r<1, so the series will converge and have a sum
S=t_1/(1-r)
t_1=64
S=64/(1/4)=256
Write the first 8 rows of Pascal's Triangle
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
pg 542 # 12
Evaluate
(n!)/(3!*(n-3)!)
(n*(n-1)*(n-2)*(n-3)!)/(3*2*1*(n-3)!)=(n(n-1)(n-2))/6
pg 545 #2
Find the next term in the sequence
8, 5, 2, -1, ...
t_5=-4
d=-3
-1-3=-4
pg546 #3
For the arithmetic sequence, find t21
-4, -1.5,1, 3.5, ...
t_1=-4
d=2.5
t_n=-4+2.5(n-1)
t_21=-4+2.5(21-1)=-4+50=46
pg 547 #4
Find the eighth term of the sequence
11,-22, 44, -88, ...
r=-2
t_1=11
t_n=11(-2)^(n-1)
t_8=11(-2)^7=11(-128)=-1408
pg 546 #8
Write the series using sigma notation
47+41+35+...+5
d=-6
t_1=47
t_n=47-6(n-1)=53-6n
\sum_(n=1)^(8)(53-6n)
Or
\sum_(n=0)^(7)(47-6n)
pg546 #8
Find the sum of the geometric series
\sum_(n=1)^(5)2ยท3^(n-1)
t_1=2, r=3,n=5
S_n=(t_1(1-r^n))/(1-r)
S_5=(2(1-3^5))/(-2)=(2(-242))/(-2)=242
pg 546 #9
Find the sum of the infinite geometric series if it has one.
250+150+90+54+...
r=150/250=3/5
r<1 so the series will converge and has a sum
t_1=250
S=t_1/(1-r)
S=250/(2/5)=625
pg 546 #12
Expand and simplify
(x^2-2)^4
The fourth row of Pascal's triangle is
1 4 6 4 1
a=x^2,b=-2
(x^2)^4+4(x^2)^3(-2)+6(x^2)^2(-2)^2+4(x^2)(-2)^3+(-2)^4
x^8-8x^6+24x^4-32x^2+16
pg546 #11
Find the 12th term in the expansion of
(2x-y)^13
the 12th term in the expansion is
k=11, a=2x, b=-y
(13!)/(2!*11!)a^1b^(12)
((13)(12))/(2*1)*2x*(-y)^12
78xy^12
pg 546 #1
Tell whether the sequence is arithmetic, geometric, or neither. Then find the next term in the sequence
2/3, 3/5, 4/7, 5/9,...
neither
t_5=6/11
pg. 547 #3
Insert three arithmetic means between 1 and 47
t_1=1
t_5=47=1+d(4)
46=4d
d=11.5
t_2=1+11.5(1)=12.5
t_3=1+11.5(2)=24
t_4=1+11.5(3)=35.5
pg 546 #5
For the geometric sequence, find the 9th term
81/8,27/4,9/2,3,...
r=3/(9/2)=2/3
t_1=81/8
t_n=81/8(2/3)^(n-1)
t_9=81/8(2/3)^8=81/8(256/6561)=32/81
pg 547 #6
Write the series in summation notation
1/2-1/4+1/6-1/8+...
t_1=1/2
Even numbers on the bottom, so the denominator is 2n. Top alternates from 1 to -1
\sum_(n=1)^(\infty)(-1)^(n+1)/(2n)
pg 546 #7
Find the sum of the first 20 terms of the arithmetic series
53+46+39+32+...
t_1=53, d=-7, n=20,t_20=-80
S_n=(n(t_1+t_n))/2
S_20=(20(53-80))/2=-270
pg 533 #12
For each geometric series, find the sum. If the series has no sum, say so.
\sum_(n=1)^(\infty)(2^n)/(5^n)
t_1=2/5
t_2=4/25
r=(4/25)/(2/5)=(2/5)
r<1 so the series will converge and have a sum
S=t_1/(1-r)
S=(2/5)/(3/5)=2/3
pg 546 #10
Expand and simplify
(a^2+b)^8
Row 8 of Pascal's triangle is 1, 8, 28, 56, 70, 56, 28, 8, 1
a^2=a, b=b
(a^2)^8+8((a^2)^7b)+28(a^2)^6b^2+56(a^2)^5b^3+70(a^2)^4b^4+56(a^2)^3b^5+28(a^2)^2b^6+8a^2b^7+b^8
a^16+8a^14b+28a^12b^2+56a^10b^3+70a^8b^4+56a^6b^5+28a^4b^6+8a^2b^7+b^8
pg546 #13
Find the fourth term in the expansion of
(3x-y^2)^6
the fourth term of the expansion is
(6!)/((6-3)!3!)a^(6-3)b^3
a=3x
b=-y^2
(6*5*4)/(3*2*1)(3x)^(3)(-y^2)^3
(120)/6*27x^3*(-y^6)
-540x^3y^6