1. ## hw help

K i think i got it thanks

First question
Write an expression for the apparent
nth term of the sequence as a function of n:
-
1, 2, 7, 13, 23, …

So yeah, this one’s hard, I’ll have to think about it

Find a formula for
nth term of the arithmetic sequence as a function of n, where
a5=190
and a10=115

The terms of an arithmetic sequence are given by:

a_n = a_1 + (n – 1)d, where a_n is the nth term, a_1 is the first term, n is the current number of term, and d is the common difference

we are told:

a_5 = a_1 + (5 – 1)d = 190
that is, a_1 + 4d = 190

a_10 = a_1 + (10 – 1)d = 115
that is, a_1 + 9d = 115

So we obtain the simultaneous equations:
a_1 + 4d = 190 ……………….(1)
a_1 + 9d = 115 ………………..(2)

=> 5d = -75 ...............................(3) = (2) - (1)
=> d = -15

But a_1 + 4d = 190
=> a_1 -4*15 = 190
=> a_1 = 190 + 4*15
=> a_1 = 250

So the terms of this arithmetic series are given by:
a_n = 250 + (n – 1)(-15)
=> a_n = 250 + 15 – 15n
=> a_n = 265 – 15n

Third question

Use the concepts of geometric sequences to write 1.3245 (i.e., 1.324545454545…)
as rational number in the form a/b

where
a and b are integers.
1.32454545… = 1 + 32/100 + 4/10^3 + 5/10^4 + 4/10^5 + 5/10^6 + …
………………= 1 + 32/100 + 45/10^4 + 45/10^6 + 45/10^8 + …
………………= 1 + 32/100 + (45/10^4)(1 + 1/10^2 + 1/10^4 + 1/10^6 + …)
………………= 1 + 32/100 + (45/10^4)(1 + (1/10^2) + (1/10^2)^2 + (1/10^2)^3 + …)
………………= 1 + 32/100 + (45/10^4)(1 + (1/100) + (1/100)^2 + (1/100)^3 + …)

But 1 + (1/100) + (1/100)^2 + (1/100)^3 + … is a geometric series with a_1 = 1 and r = 1/100. Thus it’s infinite sum is given by 1/(1 – 1/100) = 100/99.

So 1.32454545… = 1 + 32/100 + (45/10^4)(100/99)
………………….= 1 + 32/100 + (45/10^2)(1/99)
………………….= 1 + 32/100 + 45/9900
………………….= 1457/1100

Fourth question

Let a1=0
a2=1 and ak + 1 = ak + ak - 1 for k ³ 2. Write the first 8 terms of this sequence.
a_1 = 0
a_2 = 1
a_3 = a_(2+1) = a_2 + a_1 = 1 + 0 = 1
a_4 = a_(3+1) = a_3 + a_2 = 1 + 1 = 2
a_5 = a_(4+1) = a_4 + a_3 = 1 + 2 = 3
a_6 = a_(5+1) = a_5 + a_4 = 2 + 3 = 5
a_7 = a_(6+1) = a_6 + a_5 = 5 + 3 = 8
a_8 = a_(7+1) = a_7 + a_6 = 8 + 5 = 13

So the first 8 terms are 0, 1, 1, 2, 3, 5, 8, 13

Fifth question

How much money will I have been paid over my
30 year career if my starting salary is $40,000, and I receive a$1,000 salary raise for each year I work
So you get $1000 raise for every year you work, so the total money you get from raises is 30*1000 = 30000 Now you have 40000 base pay, so after 30 years that becomes 30*40000 = 1200000 So your total pay is 30000 + 1200000 = 1230000 So over your 30 year career you were paid$1230000, not bad. Well, it wouldn’t be bad if you weren’t in this economy