calculus

Find the sum of the series:
10 + 14 + 18 + 22 +...+138
and 6-12+24-48+96-...+1536

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  1. The first one is an arithmetic series where a=10 and d=4
    Use 138 as the last term to find the number of terms in the series, then use the sum of terms formula.

    The second is a geometric series with a=6 and r=-2
    Use 1536 to find the number of terms you have, then the sum of terms of a GS formula

    Surely since you are studying this topic you must have the formulas required to answer the above.
    Let me know what you got.

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  2. 10 + 14 + 18 + 22 +...+138

    arithmetic series
    difference = d = 4
    sum = (n/2)(a1 + an)
    where
    138 = 10 + 4(n-1)
    128 = 4 n - 4
    132 = 4 n
    n = 33
    so
    sum = (33/2)(148)
    =2442

    geometric
    6-12+24-48+96-...+1536

    6 + (-2)6 + (-2)^2 (6) + (-2)^3 (6)
    g = 6
    r = -2

    last term is g r^(n-1)
    1536 = 6 (-2)^n-1)
    256 = (-2)^(n-1)
    log 256 = (n-1) log (-2)
    well, lets hope n-1 is even so -2^n = 2^n. it is +1536 so it is OK
    2.40824 = (n-1) .30103
    n - 1 = 8
    n = 9
    so sum = 6(1- (-2)^9 )/ (1-(-2))
    = 6(1+512)/3
    =1026

    An+1 = -2 An

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  3. ok for the first one i got stuck at:
    n^2 + 4n - 69=0
    now factoring doesn't work. what should i do??

    second one
    i get stuck at -4608=6(-2^n - 1)

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