Find the sum of the series:

10 + 14 + 18 + 22 +...+138
and 6-12+24-48+96-...+1536

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.

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

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)

To find the sum of a series, we first need to determine its pattern and then use a formula to calculate the sum.

For the series 10 + 14 + 18 + 22 + ... + 138, we can see that each term increases by 4. This is an arithmetic series with a common difference of 4.

To find the sum of an arithmetic series, we use the formula:
Sum = (n/2)(first term + last term)
where n is the number of terms.

First, we need to find the number of terms in the series. The first term is 10, and the last term is 138. To find the number of terms, we can use the formula:
last term = first term + (n-1) * common difference

138 = 10 + (n-1) * 4
128 = (n-1) * 4
32 = n-1
n = 33

Now that we have the number of terms (33), we can substitute this value into the sum formula to find the sum:
Sum = (33/2)(10 + 138)
Sum = (33/2)(148)
Sum = 33 * 74
Sum = 2442

Therefore, the sum of the series 10 + 14 + 18 + 22 + ... + 138 is 2442.

For the series 6 - 12 + 24 - 48 + 96 - ... + 1536, we can see that each term alternates between multiplying by -2 and multiplying by 2. This is a geometric series with a common ratio of -2 and 2.

To find the sum of a geometric series, we use the formula:
Sum = (first term * (1 - common ratio^n)) / (1 - common ratio)

First, let's find the number of terms. In this series, each term is being multiplied by -2 and then by 2 alternatively. So there are two terms for each power of 2. We need to find the largest power of 2 that is less than or equal to 1536.

2^9 = 512
2^10 = 1024
2^11 = 2048

Since 1536 is less than 2048 but greater than 1024, we can take 2^10 as the largest power of 2. This means there are 10 terms.

Now we can substitute the values into the sum formula to find the sum:
Sum = (6 * (1 - (-2)^10)) / (1 - (-2))
Sum = (6 * (1 - 1024)) / (1 + 2)
Sum = (6 * (-1023)) / 3
Sum = -2046

Therefore, the sum of the series 6 - 12 + 24 - 48 + 96 - ... + 1536 is -2046.