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Math Help - Sequnces and Series help(and on limit question)

  1. #1
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    Sequnces and Series help(and on limit question)

    I need help with these four off my review sheet. I would love some explanations

    1. Find the specified term of the arithmetic sequence:
    t2=9, t5=21; 41=__

    2. Find the specified term of the geometric sequence:
    t2=8,t4=72; find t6=__

    3. Find the sum of the arithmetic series. If there is no sum, state so.
    50+48+46+ ... +10

    4. Sorry im not sure how to type this question so try to bare with me lol
    Lim x-->infinity (radical(1+x)-1)/(x)
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  2. #2
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Porcelain View Post
    I need help with these four off my review sheet. I would love some explanations

    1. Find the specified term of the arithmetic sequence:
    t2=9, t5=21; 41=_
    In one arithmetic sequence we have : next term = previous term + number
    In your example we will have...
    t1+x=t2
    t2=9
    t3=t2+x
    t4=t3+x
    t5=t4+x
    t5=21

    By doing the replacements we get :
    21=t4+x \Rightarrow 21=t3+x+x \Rightarrow 21=t2+x+x+x \Rightarrow 21=9+3x \Rightarrow 3x=12 \Rightarrow x=4
    So we have t1=5 , t2=9 , t3=13 , t4=17 , t5=21 , t6=25 .... and we can find that t10=41


    Quote Originally Posted by Porcelain View Post
    2. Find the specified term of the geometric sequence:
    t2=8,t4=72; find t6=__
    In geometric sequence we have : next term=previous term*number (say y). Here we have :
    t2=8
    t3=8*y
    t4=t3*y
    Working like before we find y^2=9 \Rightarrow y=3 or y=-3

    Now you can find t6.

    Hope this is what you want...
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  3. #3
    Member SENTINEL4's Avatar
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    Quote Originally Posted by Porcelain View Post
    3. Find the sum of the arithmetic series. If there is no sum, state so.
    50+48+46+ ... +10
    You can find the sum from
    \sum=\frac{(a1+an)*n}{2} , where a1=first term, an=last term, n=number of terms
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  4. #4
    No one in Particular VonNemo19's Avatar
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    Check this out

    \lim_{x\toinfty}\frac{\sqrt{1+x}-1}{x}=\lim_{x\toinfty}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}=\lim_{x\toinfty  }\frac{x}{x(\sqrt{1+x}-1)}=

    =\lim_{x\toinfty}\frac{1}{\sqrt{1+x}+1}

    You can see that the function is only defined for x\geq{-1} and that as x increases without bound, it is implied that f(x)\rightarrow0.

    SORRY ABOUT THE ERROR GUYS!
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  5. #5
    No one in Particular VonNemo19's Avatar
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    Check this out

    \lim_{x\to\infty}\frac{\sqrt{1+x}-1}{x}=\lim_{x\to\infty}\frac{(\sqrt{1+x}-1)(\sqrt{1+x}+1)}{x(\sqrt{1+x}+1)}=\lim_{x\to\inft  y}\frac{x}{x(\sqrt{1+x}-1)}=

    =\lim_{x\to\infty}\frac{1}{\sqrt{1+x}+1}

    You can see that the function is only defined for x\geq{-1} and that as x increases without bound, it is implied that f(x)\rightarrow0
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  6. #6
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    Quote Originally Posted by Porcelain View Post
    3. Find the sum of the arithmetic series. If there is no sum, state so.
    50+48+46+ ... +10
    Recall that \sum\limits_{k = 1}^n k  = \left[ {\frac{{n(n + 1)}}{2}} \right].

    \sum\limits_{k = 5}^{25} {2k}  = 2\sum\limits_{k = 5}^{25} k  = 2\left[ {\sum\limits_{k = 1}^{25} k  - \sum\limits_{k = 1}^4 k } \right] = 2\left[ {\frac{{25(26)}}{2} - \frac{{4(5)}}{2}} \right]
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  7. #7
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    thanks everyone i think this will be my first 100 in this class
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