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  1. #1
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    hw help

    K i think i got it thanks
    Last edited by ladventchildp; April 15th 2007 at 02:31 AM.
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by ladventchildp View Post
    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 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
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